Maxima, Minima and Mixed Practice — Worked Solutions

Ques­tion 1: Local max­ima and min­ima of two func­tions#

The prob­lem#

Find local max­ima and min­ima: (a) $f(x) = x^3 - 3x^2 - 9x + 4$; (b) $\displaystyle f(x) = x + \frac{4}{x}$ for $x > 0$.

Under­stand­ing the prob­lem#

A local max­i­mum is a point higher than all nearby points; a local min­i­mum is lower than all nearby points. You must find where they occur and the val­ues of $f$ there.

The idea#

Find the crit­i­cal points by solv­ing $f'(x) = 0$, then decide the nature of each with the sec­ond deriv­a­tive test: $f'' < 0$ gives a max­i­mum, $f'' > 0$ a min­i­mum.

Step-by-step solu­tion#

Part (a)

Step 1. Dif­fer­en­ti­ate and fac­torise.

$$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$$

Step 2. Crit­i­cal points: $x = 3$ and $x = -1$.

Step 3. Sec­ond deriv­a­tive: $f''(x) = 6x - 6$.

$$f''(-1) = -12 < 0 \;(\text{maximum}), \qquad f''(3) = 12 > 0 \;(\text{minimum})$$

Step 4. Val­ues.

$$\begin{aligned} f(-1) &= -1 - 3 + 9 + 4 = 9 \\ f(3) &= 27 - 27 - 27 + 4 = -23 \end{aligned}$$

Part (b)

Step 1. Dif­fer­en­ti­ate.

$$\displaystyle f'(x) = 1 - \frac{4}{x^2}$$

Step 2. Set $f'(x) = 0$: $x^2 = 4$, and since $x > 0$, $x = 2$.

Step 3. Sec­ond deriv­a­tive: $\displaystyle f''(x) = \frac{8}{x^3}$, so $f''(2) = 1 > 0$: a local min­i­mum.

Step 4. Value: $\displaystyle f(2) = 2 + \tfrac42 = 4$.

Check­ing the answer#

(a) Near $x = -1$: $f(-1.1) \approx 8.94$ and $f(-0.9) \approx 8.94$, both below $9$. ✓
(b) $f(1) = 5$ and $f(4) = 5$, both above $4$. ✓

Answer#

(a) Local max­i­mum $9$ at $x = -1$; local min­i­mum $-23$ at $x = 3$. (b) Local min­i­mum $4$ at $x = 2$ (no local max­i­mum).

Ques­tion 2: Absolute extremes of a quar­tic on an inter­val#

The prob­lem#

Find absolute extremes of $f(x) = x^4 - 8x^2 + 3$ on $[-1, 3]$.

Under­stand­ing the prob­lem#

On a closed inter­val a con­tin­u­ous func­tion has a great­est and a least value. They occur either at a crit­i­cal point inside the inter­val or at an end­point.

The idea#

Find crit­i­cal points, keep only those in $[-1, 3]$, then eval­u­ate $f$ at them and at both end­points and com­pare.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate and fac­torise.

$$f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)$$

Step 2. Crit­i­cal points are $0, 2, -2$. Only $0$ and $2$ lie in $[-1, 3]$; dis­card $-2$.

Step 3. Eval­u­ate at the can­di­dates $-1, 0, 2, 3$.

$$\begin{aligned} f(-1) &= 1 - 8 + 3 = -4 \\ f(0) &= 3 \\ f(2) &= 16 - 32 + 3 = -13 \\ f(3) &= 81 - 72 + 3 = 12 \end{aligned}$$

Step 4. The largest is $12$ (at $x = 3$) and the small­est is $-13$ (at $x = 2$).

Check­ing the answer#

$f(x) = (x^2 - 4)^2 - 13$, which is never below $-13$ and equals $-13$ exactly at $x = \pm 2$. ✓

Answer#

Absolute max­i­mum $12$ at $x = 3$; absolute min­i­mum $-13$ at $x = 2$.

Com­mon mis­take to avoid#

Do not use $x = -2$: it is out­side $[-1, 3]$.

Ques­tion 3: Absolute extremes of sin 2x#

The prob­lem#

Find absolute extremes of $f(x) = \sin 2x$ on $[0, \pi]$.

Under­stand­ing the prob­lem#

This is again a closed inter­val, so com­pare val­ues at crit­i­cal points inside it and at the end­points $0$ and $\pi$.

The idea#

Solve $f'(x) = 2\cos 2x = 0$ in the inter­val.

Step-by-step solu­tion#

Step 1. $f'(x) = 2\cos 2x = 0$ means $\displaystyle 2x = \tfrac{\pi}{2}$ or $\displaystyle \tfrac{3\pi}{2}$ (since $2x \in [0, 2\pi]$), so $\displaystyle x = \tfrac{\pi}{4}$ or $\displaystyle \tfrac{3\pi}{4}$.

Step 2. Eval­u­ate.

$$\displaystyle f(0) = 0, \quad f\left(\tfrac{\pi}{4}\right) = \sin\tfrac{\pi}{2} = 1, \quad f\left(\tfrac{3\pi}{4}\right) = \sin\tfrac{3\pi}{2} = -1, \quad f(\pi) = 0$$

Step 3. Com­pare: largest $1$, small­est $-1$.

Check­ing the answer#

$\sin$ of any­thing lies between $-1$ and $1$, so these are the best pos­si­ble val­ues. ✓

Answer#

Absolute max­i­mum $1$ at $\displaystyle x = \tfrac{\pi}{4}$; absolute min­i­mum $-1$ at $\displaystyle x = \tfrac{3\pi}{4}$.

Ques­tion 4: x³ has no local extreme#

The prob­lem#

Show that $f(x) = x^3$ has no local extreme.

Under­stand­ing the prob­lem#

This is a proof. A local extreme of a dif­fer­en­tiable func­tion can occur only at a crit­i­cal point, so you must look at every crit­i­cal point and show it is not an extreme.

The idea#

Use the first deriv­a­tive test: an extreme needs $f'$ to change sign.

Step-by-step solu­tion#

Step 1. $f'(x) = 3x^2$, which is $0$ only at $x = 0$. So $x = 0$ is the only can­di­date.

Step 2. For $x < 0$, $f'(x) = 3x^2 > 0$; for $x > 0$, also $f'(x) > 0$. The sign of $f'$ does not change at $0$.

Step 3. By the first deriv­a­tive test, $x = 0$ is nei­ther a max­i­mum nor a min­i­mum (it is a point of inflec­tion). Since there is no other crit­i­cal point, $f$ has no local extreme.

Check­ing the answer#

$f(-0.1) = -0.001 < 0 = f(0) < 0.001 = f(0.1)$: there are points both below and above $f(0)$ arbi­trar­ily close to $0$. ✓

Answer#

$f'(x) = 3x^2 \ge 0$ never changes sign, so $x^3$ is always increas­ing and has no local extreme.

Com­mon mis­take to avoid#

The sec­ond deriv­a­tive test gives $f''(0) = 0$, which only says the test fails; it does not prove any­thing. Use the first deriv­a­tive test instead.

Ques­tion 5: Great­est prod­uct for a fixed sum#

The prob­lem#

Find two pos­i­tive num­bers whose sum is $30$ and whose prod­uct is great­est.

Under­stand­ing the prob­lem#

Call the num­bers $x$ and $30 - x$, with $0 < x < 30$. You must max­imise their prod­uct.

The idea#

Write the prod­uct as a func­tion of $x$ alone and dif­fer­en­ti­ate.

Step-by-step solu­tion#

Step 1. Prod­uct.

$$P(x) = x(30 - x) = 30x - x^2$$

Step 2. $P'(x) = 30 - 2x = 0$ at $x = 15$.

Step 3. $P''(x) = -2 < 0$, so this is a max­i­mum. The other num­ber is $30 - 15 = 15$.

Check­ing the answer#

$P(15) = 225$; nearby, $P(14) = 14 \times 16 = 224$ and $P(16) = 224$, both smaller. ✓

Answer#

$15$ and $15$ (prod­uct $225$).

Ques­tion 6: Rec­tan­gle of max­i­mum area from a wire#

The prob­lem#

A wire $36$ cm long is bent into a rec­tan­gle. Find the dimen­sions of max­i­mum area.

Under­stand­ing the prob­lem#

The whole wire forms the perime­ter: $2(\text{length} + \text{breadth}) = 36$, so length $+$ breadth $= 18$. You need the length and breadth that make the area largest.

The idea#

Let one side be $x$; the other is $18 - x$. Max­imise $A = x(18 - x)$.

Step-by-step solu­tion#

Step 1. Area.

$$A(x) = x(18 - x) = 18x - x^2, \qquad 0 < x < 18$$

Step 2. $A'(x) = 18 - 2x = 0$ at $x = 9$.

Step 3. $A''(x) = -2 < 0$: max­i­mum. The other side is $18 - 9 = 9$.

Check­ing the answer#

Area $81$ cm²; a $10 \times 8$ rec­tan­gle (same perime­ter) has area $80$ cm², smaller. ✓

Answer#

A square of side $9$ cm ($9 \times 9$ cm, area $81$ cm²).

Ques­tion 7: Largest rec­tan­gle in a semi­cir­cle#

The prob­lem#

A rec­tan­gle is inscribed in a semi­cir­cle of radius $5$ with one side on the diam­e­ter. Find its largest area.

Under­stand­ing the prob­lem#

Put the cen­tre of the semi­cir­cle at the ori­gin, with the diam­e­ter on the $x$-axis. The semi­cir­cle is $x^2 + y^2 = 25$, $y \ge 0$. By sym­me­try the rec­tan­gle runs from $-x$ to $x$ along the diam­e­ter, and its top cor­ners $(\pm x, y)$ lie on the arc.

The idea#

Width $= 2x$, height $= y = \sqrt{25 - x^2}$. Max­imise $A = 2x\sqrt{25 - x^2}$. Because $A > 0$, it is eas­ier to max­imise $A^2$, which is largest at the same $x$.

Step-by-step solu­tion#

Step 1. Area and its square.

$$A = 2x\sqrt{25 - x^2}, \qquad A^2 = 4x^2(25 - x^2) = 100x^2 - 4x^4, \quad 0 < x < 5$$

Step 2. Dif­fer­en­ti­ate $A^2$ with respect to $x$.

$$\displaystyle \frac{d}{dx}(A^2) = 200x - 16x^3 = 8x(25 - 2x^2)$$

Step 3. Set to zero (with $x > 0$): $\displaystyle x^2 = \tfrac{25}{2}$, so $\displaystyle x = \tfrac{5}{\sqrt2}$.

Step 4. Sign check: for $x$ slightly less, $25 - 2x^2 > 0$ (increas­ing); slightly more, $< 0$ (decreas­ing). So it is a max­i­mum.

Step 5. Area at this $x$: $\displaystyle \sqrt{25 - \tfrac{25}{2}} = \tfrac{5}{\sqrt2}$.

$$\displaystyle A = 2 \cdot \frac{5}{\sqrt2} \cdot \frac{5}{\sqrt2} = \frac{50}{2} = 25$$

Check­ing the answer#

The rec­tan­gle is $5\sqrt2 \approx 7.07$ wide and $\displaystyle \tfrac{5}{\sqrt2} \approx 3.54$ high; the semi­cir­cle's own area is $\displaystyle \tfrac{25\pi}{2} \approx 39.3$, larger than $25$, as it must be.

Answer#

The largest area is $25$ square units (width $5\sqrt2$, height $\displaystyle \tfrac{5}{\sqrt2}$).

Ques­tion 8: Open tank of least sheet area#

The prob­lem#

An open tank with square base and vol­ume $32$ m³ is to be built. Find the dimen­sions that min­imise the area of sheet used.

Under­stand­ing the prob­lem#

"Open" means there is no top. The sheet cov­ers the square base (side $x$) and four walls (each $x$ by $h$). The vol­ume $x^2h = 32$ is fixed. You must choose $x$ and $h$ to make the sheet area least.

The idea#

Use the vol­ume con­straint to write $h$ in terms of $x$, so the area is a func­tion of $x$ alone.

Step-by-step solu­tion#

Step 1. From $x^2h = 32$: $\displaystyle h = \frac{32}{x^2}$.

Step 2. Sheet area = base + four walls.

$$\displaystyle S = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$$

Step 3. Dif­fer­en­ti­ate and set to zero.

$$\displaystyle S'(x) = 2x - \frac{128}{x^2} = 0 \;\Rightarrow\; x^3 = 64 \;\Rightarrow\; x = 4$$

Step 4. $\displaystyle S''(x) = 2 + \frac{256}{x^3} > 0$: a min­i­mum.

Step 5. Height: $\displaystyle h = \frac{32}{16} = 2$.

Check­ing the answer#

Vol­ume $4 \times 4 \times 2 = 32$ ✓; sheet area $16 + 4(4)(2) = 48$ m². A base of side $3$ would need $\displaystyle 9 + \tfrac{128}{3} \approx 51.7$ m², more. ✓

Answer#

Base $4$ m $\times$ $4$ m, height $2$ m.

Ques­tion 9: Near­est points on a parabola#

The prob­lem#

Find the points on $y^2 = 4x$ near­est to $(5, 0)$.

Under­stand­ing the prob­lem#

A point on the parabola is $(x, y)$ with $y^2 = 4x$ and $x \ge 0$. You want the one whose dis­tance to $(5, 0)$ is least. Min­imis­ing the dis­tance is the same as min­imis­ing its square $D^2$, which avoids a square root.

The idea#

Write $D^2 = (x - 5)^2 + y^2$ and replace $y^2$ by $4x$.

Step-by-step solu­tion#

Step 1. Squared dis­tance as a func­tion of $x$.

$$D^2 = (x - 5)^2 + 4x = x^2 - 6x + 25$$

Step 2. Dif­fer­en­ti­ate: $2x - 6 = 0$ at $x = 3$.

Step 3. Sec­ond deriv­a­tive $2 > 0$: min­i­mum.

Step 4. The points: $y^2 = 12$, so $y = \pm 2\sqrt3$.

Check­ing the answer#

$D^2 = 9 - 18 + 25 = 16$, so $D = 4$. The ver­tex $(0, 0)$ is at dis­tance $5$, far­ther. ✓

Answer#

$(3, 2\sqrt3)$ and $(3, -2\sqrt3)$, at dis­tance $4$.

Ques­tion 10: Out­put for max­i­mum profit#

The prob­lem#

The cost of pro­duc­ing $x$ items is $C(x) = x^3 - 30x^2 + 300x + 100$ and each sells for ₹228. Find the out­put that max­imises profit.

Under­stand­ing the prob­lem#

Rev­enue from $x$ items is $228x$ rupees. Profit $=$ rev­enue $-$ cost. The num­ber of items must be a whole num­ber, so after find­ing the best $x$ you must check the near­est inte­gers.

The idea#

Form $P(x) = 228x - C(x)$, solve $P'(x) = 0$, iden­tify the max­i­mum, then com­pare neigh­bour­ing whole num­bers.

Step-by-step solu­tion#

Step 1. Profit.

$$P(x) = 228x - (x^3 - 30x^2 + 300x + 100) = -x^3 + 30x^2 - 72x - 100$$

Step 2. Dif­fer­en­ti­ate.

$$P'(x) = -3x^2 + 60x - 72 = -3(x^2 - 20x + 24)$$

Step 3. Solve $x^2 - 20x + 24 = 0$ with the qua­dratic for­mula.

$$\displaystyle x = \frac{20 \pm \sqrt{400 - 96}}{2} = 10 \pm \sqrt{76} = 10 \pm 2\sqrt{19}$$

So $x \approx 1.28$ or $x \approx 18.72$.

Step 4. Sec­ond deriv­a­tive: $P''(x) = -6x + 60$. At $x \approx 18.72$, $P'' < 0$: max­i­mum. (At $x \approx 1.28$, $P'' > 0$: that is a min­i­mum.)

Step 5. $x$ must be a whole num­ber, so com­pare $18$ and $19$.

$$\begin{aligned} P(18) &= -5832 + 9720 - 1296 - 100 = 2492 \\ P(19) &= -6859 + 10830 - 1368 - 100 = 2503 \end{aligned}$$

$P(19)$ is larger.

Check­ing the answer#

$P(20) = -8000 + 12000 - 1440 - 100 = 2460$, smaller than $P(19)$, so $19$ really is the best whole num­ber.

Answer#

Pro­duce $19$ items (max­i­mum profit ₹2503).

Com­mon mis­take to avoid#

Do not sim­ply round $18.72$ and stop — com­pare the profit at both neigh­bour­ing inte­gers.

Ques­tion 11: Largest plot against a wall#

The prob­lem#

A $40$ m fence encloses a rec­tan­gu­lar plot against a straight wall (the wall forms one side). Find the largest area.

Under­stand­ing the prob­lem#

The fence cov­ers only three sides: two sides of length $x$ (per­pen­dic­u­lar to the wall) and one side par­al­lel to the wall. So that side has length $40 - 2x$.

The idea#

Area $= x(40 - 2x)$; max­imise it for $0 < x < 20$.

Step-by-step solu­tion#

Step 1. Area.

$$A(x) = x(40 - 2x) = 40x - 2x^2$$

Step 2. $A'(x) = 40 - 4x = 0$ at $x = 10$.

Step 3. $A''(x) = -4 < 0$: max­i­mum.

Step 4. Dimen­sions $10$ m by $40 - 20 = 20$ m; area $200$ m².

Check­ing the answer#

Try $x = 9$: $9 \times 22 = 198$; $x = 11$: $11 \times 18 = 198$. Both are less than $200$. ✓

Answer#

The largest area is $200$ m² ($20$ m along the wall, $10$ m out from it).

Ques­tion 12: Local max­i­mum of x e⁻ˣ#

The prob­lem#

Find the local max­i­mum value of $f(x) = xe^{-x}$.

Under­stand­ing the prob­lem#

You need the crit­i­cal point, to con­firm it is a max­i­mum, and the value of $f$ there.

The idea#

Dif­fer­en­ti­ate with the prod­uct rule, solve $f'(x) = 0$, and use the first deriv­a­tive test.

Step-by-step solu­tion#

Step 1. Prod­uct rule.

$$f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1 - x)$$

Step 2. $e^{-x} > 0$ always, so $f'(x) = 0$ only when $x = 1$.

Step 3. Sign of $f'$: for $x < 1$, $1 - x > 0$ so $f' > 0$; for $x > 1$, $f' < 0$. The sign changes from $+$ to $-$: a local max­i­mum.

Step 4. Value.

$$\displaystyle f(1) = 1 \cdot e^{-1} = \frac1e$$

Check­ing the answer#

$\displaystyle \tfrac1e \approx 0.368$; $f(0.5) \approx 0.303$ and $f(2) \approx 0.271$, both smaller. ✓

Answer#

Local max­i­mum value $\displaystyle \frac1e$, at $x = 1$.