The highest and lowest values a function takes on a whole interval — and how to find them: list the critical points and the end points, evaluate, compare. Then the payoff: the best box, the cheapest can, the most profitable price.
A packaging engineer at a dairy co-operative must design a one-litre can that uses as little metal as possible; a farmer with $240$ m of fencing wants the largest paddock he can enclose against a wall; a canteen manager wants the samosa price that brings in the most profit. In each case one quantity — metal, area, profit — depends on a choice we are free to make, and we want the best possible value, not merely a value better than its neighbours. These are optimisation problems.
Section 6.3 found local maxima and minima — peaks and valleys compared with nearby points. Here we want the absolute maximum and minimum: the largest and smallest values over the whole interval of allowed choices. On a closed interval the method is short and reliable: the answer is at a critical point or at an end point, so we list them, evaluate the function at each, and compare. Then we put the method to work on real design problems.
Absolute Maximum and Minimum on a Closed Interval
📘 Definition 6.7 — Absolute maximum and minimum
Let $f$ be defined on a set $I$ (usually an interval) and let $c\in I$.
$f(c)$ is the absolute maximum value of $f$ on $I$ if $f(c)\ge f(x)$ for every $x\in I$;
$f(c)$ is the absolute minimum value of $f$ on $I$ if $f(c)\le f(x)$ for every $x\in I$.
The point $c$ is then a point of absolute maximum (minimum). Together these are the absolute extreme values of $f$ on $I$; they are also called the greatest and least values.
A function need not have them. On the open interval $(0,1)$, $f(x)=x$ takes values as close to $1$ as you like but never $1$, so it has no absolute maximum (and no absolute minimum either). On all of $\mathbb R$, $x^3$ has neither. The trouble comes from missing end points or unbounded intervals. On a closed, bounded interval it cannot happen to a continuous function:
📐 Theorem 6.6 — Continuous functions on closed intervals
If $f$ is continuous on the closed interval $[a,b]$, then $f$ has an absolute maximum value and an absolute minimum value on $[a,b]$: there are points $c,d\in[a,b]$ with $f(d)\le f(x)\le f(c)$ for every $x\in[a,b]$.
We accept Theorem 6.6 without proof: it depends on the completeness of the real numbers, which is studied at university. It says that the extremes exist; the next theorem says where to look for them.
📐 Theorem 6.7 — Where absolute extrema occur
Let $f$ be continuous on $[a,b]$. Then each absolute extreme value of $f$ on $[a,b]$ is attained either at an end point $a$ or $b$, or at a critical point of $f$ in the open interval $(a,b)$.
Why. If the absolute maximum is attained at a point $c$ strictly inside $(a,b)$, then $f(c)\ge f(x)$ for all $x$ near $c$ in particular, so $c$ is a point of local maximum, and by Theorem 6.3 either $f'(c)=0$ or $f'(c)$ does not exist — $c$ is a critical point. The same argument works for the minimum. $\square$
🧭 Problem-solving strategy — absolute extrema on $[a,b]$
Check that $f$ is continuous on $[a,b]$.
Find all critical points of $f$ that lie inside $(a,b)$: solve $f'(x)=0$ and note points where $f'$ does not exist. Discard those outside the interval.
Make a table of $f$ at every such critical point and at both end points.
The largest entry is the absolute maximum value, the smallest is the absolute minimum value. No first or second derivative test is needed.
✏️ Example 1 — A cubic on a closed interval
Find the absolute maximum and minimum values of $f(x)=2x^3-15x^2+36x+1$ on $[1,5]$.
💡 Show solution▶
Critical points $f'(x)=6x^2-30x+36=6(x-2)(x-3)$; both $2$ and $3$ lie in $(1,5)$.
Table
$x$
$1$ (end)
$2$
$3$
$5$ (end)
$f(x)$
$24$
$29$
$28$
$56$
Answer Absolute maximum value $56$ at $x=5$; absolute minimum value $24$ at $x=1$.
Lesson Both extremes are at end points here, and the local maximum $f(2)=29$ is not the absolute maximum. Never skip the end points.
✏️ Example 2 — A trigonometric function
Find the absolute maximum and minimum values of $f(x)=\sin x+\cos x$ on $[0,\pi]$.
💡 Show solution▶
Critical points $f'(x)=\cos x-\sin x=0\iff\tan x=1$; in $(0,\pi)$ only $x=\tfrac\pi4$.
Answer Absolute maximum $\sqrt2\approx1.414$ at $x=\tfrac\pi4$; absolute minimum $-1$ at $x=\pi$.
✏️ Example 3 — A critical point where $f'$ does not exist
Find the absolute maximum and minimum values of $f(x)=|x^2-4|$ on $[-3,1]$.
💡 Show solution▶
Pieces $f(x)=x^2-4$ when $|x|\ge2$ and $f(x)=4-x^2$ when $|x|\lt2$. $f$ is continuous (absolute value of a continuous function).
Critical points On $(-3,-2)$: $f'=2x\ne0$. On $(-2,1)$: $f'=-2x=0$ at $x=0$. At $x=-2$ the left derivative is $2(-2)=-4$ and the right derivative is $-2(-2)=4$: a corner, $f'(-2)$ does not exist. Critical points in $(-3,1)$: $x=-2$ and $x=0$.
Table $f(-3)=5$, $f(-2)=0$, $f(0)=4$, $f(1)=3$.
Answer Absolute maximum $5$ at $x=-3$; absolute minimum $0$ at $x=-2$ — at the corner, which solving $f'(x)=0$ alone would have missed.
Interactive 1 — Absolute Extrema on a Closed Interval
Choose a function and set the end points $a$ and $b$ (Play slides the interval along). The lab lists the candidates — both end points and every critical point inside $(a,b)$ — with their values, and marks the absolute maximum and absolute minimum. Watch the winner jump between a critical point and an end point as the interval moves.
interval—
candidates—
absolute max—
absolute min—
Optimisation Problems
In applications the interval of allowed choices is often open — a box cannot have a cut of size $0$ or of half the sheet, a can cannot have radius $0$. Then Theorem 6.7 does not apply directly, but the following fact almost always finishes the job.
📐 Theorem 6.8 — A single critical point
Let $f$ be continuous on an interval $I$ (of any kind) and suppose $f$ has exactly one critical point $c$ in the interior of $I$. If $c$ is a point of local maximum, then $f(c)$ is the absolute maximum value of $f$ on $I$; if $c$ is a point of local minimum, then $f(c)$ is the absolute minimum value of $f$ on $I$.
Why. Say $c$ is a local maximum. On each side of $c$ there is no other critical point, so $f'$ keeps one sign there (Section 6.2). Just left of $c$ it is positive and just right of $c$ negative (first derivative test), so $f$ increases all the way up to $c$ and decreases all the way after it: no point of $I$ can beat $f(c)$. $\square$
🧭 Problem-solving strategy — optimisation
Understand and draw. Sketch the object; label every length with a letter.
Name the target. Which quantity is to be made largest or smallest? Write a formula for it.
Use the condition (fixed perimeter, volume, material, budget) to express the target in terms of one variable.
State the domain of that variable from the physical situation (lengths positive, cuts smaller than the sheet, …).
Differentiate, find the critical points in the domain, and justify the maximum or minimum: closed-interval table, or Theorem 6.8 with the first or second derivative test.
Answer the question asked — dimensions, the best value, or both — with units, and check that it makes sense.
✏️ Example 4 — The open box from a rectangular sheet
An open box is made from a rectangular sheet of cardboard $30$ cm by $16$ cm by cutting equal squares of side $x$ cm from the four corners and folding up the sides. Find the value of $x$ that gives the box of largest volume, and the largest volume.
💡 Show solution▶
Target The base is $(30-2x)$ by $(16-2x)$ and the height is $x$, so $V(x)=x(30-2x)(16-2x)$. The sides must be positive: $0\lt x\lt8$.
Expand and differentiate $V(x)=4x^3-92x^2+480x$, so $V'(x)=12x^2-184x+480$ $=4(3x^2-46x+120)$ $=4(3x-10)(x-12)$.
Critical point $x=\tfrac{10}3$ or $x=12$; only $x=\tfrac{10}3$ lies in $(0,8)$. $V''(x)=24x-184$, and $V''\left(\tfrac{10}3\right)=80-184=-104\lt0$: a local maximum, hence by Theorem 6.8 the absolute maximum on $(0,8)$.
Answer Cut squares of side $\tfrac{10}3\approx3.33$ cm. The box is $\tfrac{70}3$ cm by $\tfrac{28}3$ cm by $\tfrac{10}3$ cm, with volume $\dfrac{10\cdot70\cdot28}{27}=\dfrac{19600}{27}\approx725.9\ \text{cm}^3$.
Check $V(3)=3\cdot24\cdot10=720$ and $V(3.5)=3.5\cdot23\cdot9=724.5$, both a little less ✓.
✏️ Example 5 — The cheapest one-litre can
A closed cylindrical can must hold $1$ litre $=1000\ \text{cm}^3$. Find the radius and height that use the least metal (least total surface area).
💡 Show solution▶
Target and condition Surface area $S=2\pi r^2+2\pi rh$ (two ends and the curved side); volume $\pi r^2h=1000$, so $h=\dfrac{1000}{\pi r^2}$ and $S(r)=2\pi r^2+\dfrac{2000}{r}$, $r\gt0$.
Critical point $S'(r)=4\pi r-\dfrac{2000}{r^2}=0\iff r^3=\dfrac{500}\pi$, so $r=\left(\dfrac{500}{\pi}\right)^{1/3}\approx5.42$ cm.
Minimum $S''(r)=4\pi+\dfrac{4000}{r^3}\gt0$ for all $r\gt0$: a local minimum, and the only critical point, so the absolute minimum (Theorem 6.8).
Answer $h=\dfrac{1000}{\pi r^2}=\dfrac{2\cdot500}{\pi r^3}\cdot r=2r\approx10.84$ cm. The best can is as tall as it is wide ($h=2r$), with $S=6\pi r^2\approx553.6\ \text{cm}^2$.
Reality check Real cans are usually taller than this. Their ends are made of thicker metal and the seams cost extra, which shifts the best shape — a nice question for the mini project.
✏️ Example 6 — Fencing against a wall
A farmer has $240$ m of fencing to make a rectangular paddock against a long straight wall; the wall forms one side and needs no fence. Find the dimensions of the largest paddock.
💡 Show solution▶
Target Let the two sides perpendicular to the wall be $x$ m each; the side parallel to the wall is $240-2x$ m. Area $A(x)=x(240-2x)=240x-2x^2$, with $0\le x\le120$.
Closed interval $A'(x)=240-4x=0$ at $x=60$. Table: $A(0)=0$, $A(60)=7200$, $A(120)=0$.
Answer $60$ m by $120$ m (the long side along the wall); area $7200\ \text{m}^2$.
Compare A square of side $80$ m (using $240$ m on three sides) gives only $6400\ \text{m}^2$. The wall is free fencing, so the best shape uses more of it.
Interactive 2 — Optimisation Lab
Choose a design problem. Move the slider (or press Play) to change the design: the picture on the left changes, and the right panel plots the quantity being optimised. The best design — where the derivative is zero — is marked on the graph. Can you beat it?
choice—
quantity—
derivative—
best possible—
⚠️ Common mistakes when studying alone
1. Forgetting the end points. On a closed interval the absolute extreme is often at an end (Example 1). Always put $f(a)$ and $f(b)$ in the table.
2. Using critical points outside the interval. In Example 4, $x=12$ solves $V'=0$ but is impossible for a $16$ cm wide sheet. State the domain first.
3. Two variables left in the formula. $S=2\pi r^2+2\pi rh$ cannot be optimised until the condition $\pi r^2h=1000$ removes $h$.
4. No justification. "$V'=0$ at $x=\tfrac{10}3$, so this is the maximum" loses marks. Say why: a table, the second derivative, or Theorem 6.8.
5. Answering a different question. If the question asks for the dimensions, give all of them; if it asks for the largest volume, give the value with units.
Exercises
I
Find the absolute maximum and absolute minimum values on the given interval.
1 $f(x)=x^2-4x+1$ on $[0,5]$
2 $f(x)=4x-\tfrac12x^2$ on $\left[-2,\tfrac92\right]$
3 $f(x)=x^3-3x$ on $[0,2]$
4 $f(x)=(x-1)^2+3$ on $[-3,1]$
5 $f(x)=x+2\cos x$ on $[0,\pi]$
6 $f(x)=xe^{-x}$ on $[0,3]$
7 $f(x)=|x-2|+1$ on $[0,5]$
8 $f(x)=x^4-2x^2+3$ on $[-2,1]$
9 $f(x)=x-2\sqrt x$ on $[0,9]$
💡 Show solutions▶
1 $f'=2x-4$, critical point $2$. $f(0)=1$, $f(2)=-3$, $f(5)=6$. Max $6$ at $x=5$; min $-3$ at $x=2$.
2 $f'=4-x$, critical point $4$. $f(-2)=-10$, $f(4)=8$, $f\left(\tfrac92\right)=7.875$. Max $8$ at $x=4$; min $-10$ at $x=-2$.
3 $f'=3x^2-3$, critical point $1$ in $(0,2)$. $f(0)=0$, $f(1)=-2$, $f(2)=2$. Max $2$ at $x=2$; min $-2$ at $x=1$.
4 $f'=2(x-1)$ has no zero inside $(-3,1)$. $f(-3)=19$, $f(1)=3$. Max $19$ at $x=-3$; min $3$ at $x=1$.
5 $f'=1-2\sin x=0$ at $x=\tfrac\pi6,\tfrac{5\pi}6$. $f(0)=2$, $f\left(\tfrac\pi6\right)=\tfrac\pi6+\sqrt3\approx2.256$, $f\left(\tfrac{5\pi}6\right)=\tfrac{5\pi}6-\sqrt3\approx0.886$, $f(\pi)=\pi-2\approx1.142$. Max $\tfrac\pi6+\sqrt3$; min $\tfrac{5\pi}6-\sqrt3$.
6 $f'=(1-x)e^{-x}$, critical point $1$. $f(0)=0$, $f(1)=\tfrac1e\approx0.368$, $f(3)=3e^{-3}\approx0.149$. Max $\tfrac1e$ at $x=1$; min $0$ at $x=0$.
7 Corner at $x=2$ ($f'$ does not exist). $f(0)=3$, $f(2)=1$, $f(5)=4$. Max $4$ at $x=5$; min $1$ at $x=2$.
8 $f'=4x(x-1)(x+1)$; critical points in $(-2,1)$: $-1$, $0$. $f(-2)=11$, $f(-1)=2$, $f(0)=3$, $f(1)=2$. Max $11$ at $x=-2$; min $2$ at $x=-1$ and at $x=1$.
9 $f'=1-\dfrac1{\sqrt x}$ ($x\gt0$) is $0$ at $x=1$. $f(0)=0$, $f(1)=-1$, $f(9)=3$. Max $3$ at $x=9$; min $-1$ at $x=1$.
II
Optimisation in geometry.
1 Find two positive numbers whose sum is $24$ and whose product is as large as possible.
2 Find two positive numbers whose product is $64$ and whose sum is as small as possible.
3 Of all rectangles with perimeter $60$ m, find the one with the largest area.
4 A $12$ cm square sheet of tin has equal squares cut from its corners and the sides folded up to make an open box. Find the largest possible volume.
5 A closed cylindrical tank is to hold $250\pi\ \text{m}^3$. Find the radius and height that minimise its total surface area, and that area.
6 An open-top cylindrical tank must hold $8\pi\ \text{m}^3$. Find the dimensions that use the least sheet metal (base and curved side).
7 Find the point on the curve $y=\sqrt x$ nearest to the point $(3,0)$, and the least distance.
8 A rectangle is inscribed in a semicircle of radius $10$ cm, with one side on the diameter. Find the largest possible area.
9 A poster must have $384\ \text{cm}^2$ of printed area, with margins of $3$ cm at top and bottom and $2$ cm at each side. Find the dimensions of the poster that use the least paper.
💡 Show solutions▶
1 $P(x)=x(24-x)$, $P'=24-2x=0$ at $x=12$, $P''=-2\lt0$: $12$ and $12$ (product $144$).
2 $S(x)=x+\dfrac{64}x$, $x\gt0$; $S'=1-\dfrac{64}{x^2}=0$ at $x=8$, $S''=\dfrac{128}{x^3}\gt0$: $8$ and $8$ (sum $16$).
3 Sides $x$ and $30-x$: $A=x(30-x)$, maximum at $x=15$. The square $15\times15$, area $225\ \text{m}^2$.
4 $V=x(12-2x)^2$, $0\lt x\lt6$; $V'=(12-2x)(12-6x)=0$ at $x=2$ (as $x=6$ is excluded). $V''(2)=-48\lt0$. Largest volume $V(2)=2\cdot8^2=128\ \text{cm}^3$.
5 $h=\dfrac{250}{r^2}$, $S=2\pi r^2+\dfrac{500\pi}r$; $S'=4\pi r-\dfrac{500\pi}{r^2}=0$ at $r^3=125$, $r=5$ m, $h=10$ m ($=2r$ again). $S=50\pi+100\pi=150\pi\approx471.2\ \text{m}^2$.
6 $h=\dfrac8{r^2}$, $S=\pi r^2+2\pi rh=\pi r^2+\dfrac{16\pi}r$; $S'=2\pi r-\dfrac{16\pi}{r^2}=0$ at $r^3=8$: $r=2$ m, $h=2$ m (an open tank has $h=r$). $S=12\pi\approx37.7\ \text{m}^2$; $S''\gt0$ confirms the minimum.
9 Printed $w\times\dfrac{384}w$; paper $A(w)=(w+4)\left(\dfrac{384}w+6\right)$ $=408+6w+\dfrac{1536}w$. $A'=6-\dfrac{1536}{w^2}=0$ at $w=16$ ($A''\gt0$). Printed $16\times24$; poster $20$ cm wide and $30$ cm tall, $600\ \text{cm}^2$.
III
Optimisation in business, transport and engineering.
1 A firm can sell $x$ units a week at a price of $p=300-2x$ rupees each; its weekly cost is $C(x)=1000+60x$ rupees. Find the output and the price that maximise weekly profit, and the maximum profit.
2 A city bus route carries $1000$ passengers a day at a fare of ₹50. Surveys suggest that every ₹1 increase in fare loses $10$ passengers a day. What fare maximises the daily revenue, and what is that revenue?
3 The strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest beam that can be cut from a round log of diameter $30$ cm.
4 A truck makes a $300$ km trip at a steady $v$ km/h. Fuel costs $\left(4+\dfrac{v^2}{1080}\right)$ rupees per km and the driver is paid ₹400 per hour. Find the speed that minimises the total cost of the trip, and that cost.
💡 Show solutions▶
1 $P(x)=x(300-2x)-(1000+60x)$ $=240x-2x^2-1000$; $P'=240-4x=0$ at $x=60$, $P''=-4\lt0$. Sell $60$ units at a price of ₹180; profit $P(60)=14400-7200-1000=6200$ rupees per week.
2 With a rise of $n$ rupees: $R(n)=(50+n)(1000-10n)$. $R'(n)=(1000-10n)-10(50+n)$ $=500-20n=0$ at $n=25$ ($R''=-20\lt0$). Fare ₹75, with $750$ passengers; revenue ₹56 250 a day (up from ₹50 000).
3 Width $w$, depth $d$, with $w^2+d^2=30^2$ (the diagonal is a diameter). Strength $S=kwd^2=kw(900-w^2)$; $S'=k(900-3w^2)=0$ at $w^2=300$: $w=10\sqrt3\approx17.3$ cm and $d=\sqrt{600}=10\sqrt6\approx24.5$ cm ($S''=-6kw\lt0$). Note $d=\sqrt2\,w$.
4 Fuel $300\left(4+\dfrac{v^2}{1080}\right)$, driver $400\cdot\dfrac{300}v$: $C(v)=1200+\dfrac{5v^2}{18}+\dfrac{120000}{v}$. $C'(v)=\dfrac{5v}9-\dfrac{120000}{v^2}=0\iff v^3=216000$, so $v=60$ km/h ($C''\gt0$). Cost $C(60)=1200+1000+2000=4200$ rupees.
IV
Reasoning.
1 Show that, of all rectangles inscribed in a circle of radius $r$, the square has the largest area.
2 Explain why $f(x)=x$ has no absolute maximum on $(0,1)$ but does on $(0,1]$. Which hypothesis of Theorem 6.6 fails in the first case?
3 Show that $-\tfrac12\le\dfrac{x}{1+x^2}\le\tfrac12$ for every real $x$, and find where the bounds are attained.
4 A function $f$ is continuous on $[0,4]$, has critical points only at $x=1$ and $x=3$, and $f(0)=2$, $f(1)=5$, $f(3)=-1$, $f(4)=3$. Find its absolute maximum and minimum values. Can you tell whether $x=3$ is a local minimum?
💡 Show solutions▶
1 Half-sides $x,y$ with $x^2+y^2=r^2$; area $A=4xy=4x\sqrt{r^2-x^2}$, $0\lt x\lt r$. $A^2=16(r^2x^2-x^4)$ has derivative $16(2r^2x-4x^3)=0$ at $x^2=\tfrac{r^2}2$, so $x=y=\tfrac r{\sqrt2}$: a square, with area $2r^2$ (the only critical point, and $A\to0$ at both ends).
2 On $(0,1)$ the values approach $1$ but $1$ is never reached, and any value $x_0$ is beaten by $\tfrac{x_0+1}2$. The interval is not closed (Theorem 6.6 needs $[a,b]$). On $(0,1]$ the maximum is $f(1)=1$; there is still no minimum.
3 $g(x)=\dfrac x{1+x^2}$, $g'(x)=\dfrac{1-x^2}{(1+x^2)^2}$: $g$ decreases on $(-\infty,-1]$, increases on $[-1,1]$, decreases on $[1,\infty)$, and $g(x)\to0$ as $x\to\pm\infty$. So the greatest value is $g(1)=\tfrac12$ and the least is $g(-1)=-\tfrac12$. (Or: $(x-1)^2\ge0$ gives $1+x^2\ge2x$.)
4 By Theorem 6.7 compare $2,5,-1,3$: absolute maximum $5$ (at $x=1$), absolute minimum $-1$ (at $x=3$). Yes: an absolute minimum at an interior point is also a local minimum.
How this is examined: Board papers ask for absolute extrema on a closed interval (2–3 marks) and nearly always have a 5-mark optimisation problem or a case study (boxes, tanks, cans, windows, prices); method marks are given for the function of one variable, the derivative, the justification and the final answer with units. JEE favours nearest-point problems and multiple-correct questions about where the extremes are attained.
E1.Board · 1 mark The absolute minimum value of $f(x)=2x^3-24x$ on $[0,3]$ is (A) $-32$ (B) $-18$ (C) $0$ (D) $2$
💡 Show solution▶
Idea $f'(x)=6x^2-24=0$ at $x=2$ in $(0,3)$. Table: $f(0)=0$, $f(2)=16-48=-32$, $f(3)=54-72=-18$.
Answer (A). Trap (B) is the value at the end point $x=3$; (D) is the critical point, not a value.
E2.Board · 1 markAssertion–Reason. Assertion (A): $f(x)=x^2$ has no absolute maximum on the open interval $(-1,1)$. Reason (R): every function continuous on a closed interval $[a,b]$ has an absolute maximum on $[a,b]$. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
💡 Show solution▶
Idea A is true: $x^2\lt1$ on $(-1,1)$, values come arbitrarily close to $1$ but never reach it. R is true (Theorem 6.6). But R is about closed intervals and says nothing about why a maximum fails on an open one.
Answer (B).
E3.Board · 5 marks An open tank with a square base and vertical sides is to hold $32\ \text{m}^3$ of water. Find the dimensions for which the area of sheet metal used (base and four sides) is least, and show that the depth is then half the width.
💡 Show solution▶
Set-up Base $x$ m by $x$ m, depth $h$ m: $x^2h=32$, so $h=\dfrac{32}{x^2}$. Area $S=x^2+4xh=x^2+\dfrac{128}x$, $x\gt0$. (2 marks)
Critical point $S'(x)=2x-\dfrac{128}{x^2}=0$ $\iff x^3=64$ $\iff x=4$. (1 mark)
Justify $S''(x)=2+\dfrac{256}{x^3}\gt0$, so $x=4$ is a local minimum and, being the only critical point, gives the least area (Theorem 6.8). (1 mark)
Answer Base $4$ m by $4$ m, depth $h=\dfrac{32}{16}=2$ m $=\tfrac12\times4$ m; least area $16+32=48\ \text{m}^2$. (1 mark)
E4.Board · Case study A school canteen sells samosas at ₹10 each and sells $300$ a day. A survey shows that every ₹1 increase in price reduces sales by $20$ a day. Each samosa costs the canteen ₹6 to make. Let the price be raised by $n$ rupees. (i) Show that the daily profit is $P(n)=(4+n)(300-20n)$. [1 mark] (ii) Find the price that gives the greatest profit. [2 marks] (iii) Find the greatest daily profit. [1 mark]
💡 Show solution▶
(i) Profit per samosa $=(10+n)-6=4+n$ rupees; number sold $=300-20n$. Their product is $P(n)$.
(ii) $P(n)=1200+220n-20n^2$, so $P'(n)=220-40n=0$ at $n=5.5$, and $P''=-40\lt0$. Price ₹15.50.
(iii) $P(5.5)=9.5\times190=1805$ rupees a day (compared with ₹1200 at the old price).
E5.JEE Main The least distance from the point $(0,2)$ to the parabola $y=x^2$ is (A) $\dfrac{\sqrt7}2$ (B) $2$ (C) $\dfrac{\sqrt3}2$ (D) $\dfrac74$
💡 Show solution▶
Method For the point $(x,x^2)$: $D^2=x^2+(x^2-2)^2=x^4-3x^2+4$. Its derivative $4x^3-6x=2x(2x^2-3)$ is zero at $x=0$ and $x=\pm\sqrt{\tfrac32}$. $D^2(0)=4$, $D^2\left(\pm\sqrt{\tfrac32}\right)=\tfrac94-\tfrac92+4=\tfrac74$.
Answer (A) $D=\sqrt{\tfrac74}=\tfrac{\sqrt7}2\approx1.32$. Trap (B): the distance to the vertex, a local maximum of $D^2$; trap (D) forgets the square root.
E6.JEE Advanced(One or more options correct.) Let $f(x)=x^3-3x^2+2$ on $[-1,3]$. Then (A) the absolute maximum value is attained at exactly two points (B) the absolute minimum value is $-2$ (C) $f$ has a local maximum at $x=0$ (D) the absolute maximum is attained at $x=2$
Options (A) The maximum $2$ is attained at $x=0$ and $x=3$ only: true. (B) The minimum is $-2$ (at $x=-1$ and $x=2$): true. (C) $f'$ goes $+\to-$ at $0$: true. (D) $x=2$ gives the minimum: false.
Answer (A), (B), (C). ⏱ Fast route: $f(x)-2=x^2(x-3)$ and $f(x)+2=(x+1)(x-2)^2$ — the repeated factors show the double values at once.
🔬 Research–Teaching NexusMini project · about 60 minutes
Research question. Are the cans and cartons in your kitchen shaped to use the least material for what they hold? Compare real packages with the optimal shape $h=2r$ from Example 5.
Why it matters. Packaging is a large share of the cost and the waste of food and drinks. Saving even a few per cent of metal or board on every can adds up to thousands of tonnes a year. But manufacturers also care about grip, stacking, labels and the thicker metal of the ends — so the "mathematically best" shape is not always what you find on the shelf.
What to do.
Collect five or more cylindrical containers (soft-drink cans, tins of tomatoes or condensed milk, a paint tin). Measure the diameter and height of each, and note the volume printed on it.
For each, compute the actual total surface area $2\pi r^2+2\pi rh$ and the ratio $\dfrac hr$.
For the same volume $V$, compute the optimal radius $r^*=\left(\dfrac V{2\pi}\right)^{1/3}$ and the least area $S^*=6\pi(r^*)^2$. Work out how much extra material, in per cent, the real container uses.
Redo the optimisation assuming the two ends cost twice as much per cm² as the side. Show that the best shape then has $h=4r$, and compare again.
Tools: Ruler or vernier calipers, spreadsheet.
Hand in: A table (container, $V$, $r$, $h$, $\tfrac hr$, $S$, $S^*$, extra %), and a short paragraph: which containers are close to optimal, which are not, and why the designers might have chosen differently.
Going further. Cuboid cartons: for a closed box with a square base and fixed volume, show that the cube uses the least board. Measure a few milk or juice cartons and compare.
Instructor note: a standard $330$ mL soft-drink can measures about $r\approx3.3$ cm, $h\approx11.5$ cm ($\tfrac hr\approx3.5$). A cylinder of those dimensions holds about $390\ \text{cm}^3$ (the printed volume leaves headspace and the ends are domed), and it uses only about $3\%$ more area than the $h=2r$ optimum for $390\ \text{cm}^3$ — but about $16\%$ more than the optimum for $330\ \text{cm}^3$; discuss which comparison is fair. With ends at twice the cost per area, cost $\propto4\pi r^2+2\pi rh$ and the optimum becomes $h=4r$, close to real drink cans. Food tins (e.g. $400$ g tomatoes) are often near $h=2r$.
🧪 Self-check quiz
0/5 correct
Q1. Which condition guarantees that $f$ has an absolute maximum and an absolute minimum on $[a,b]$?
Q2. For $f$ continuous on $[a,b]$, its absolute extreme values can occur only
Q3. The absolute maximum value of $f(x)=x^2$ on $[-1,2]$ is
Q4. Two positive numbers have sum $10$. The greatest possible value of their product is
Q5. A closed cylinder of fixed volume has the least total surface area when