🎯 Maximum and Minimum Problems

$2x - 10 = 0$, so $x = 5$.

$y = 25 - 50 + 3 = -22$

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$, so the minimum value is $-22$.

Let the sides be $x$ and $y$, with $2x + 2y = 40$, so $y = 20 - x$.

$A = x(20-x) = 20x - x^2$

$\dfrac{\mathrm{d}A}{\mathrm{d}x} = 20 - 2x = 0$, so $x = 10$.

$\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = -2 < 0$, a maximum.

$A = 10 \times 10 = 100$ cm² — the rectangle is a square.

Let them be $x$ and $20 - x$. Then $P = 20x - x^2$.

$\dfrac{\mathrm{d}P}{\mathrm{d}x} = 20 - 2x = 0$, so $x = 10$.

The numbers are both $10$, with product $100$.

$\dfrac{\mathrm{d}h}{\mathrm{d}t} = 20 - 10t = 0$, so $t = 2$ s.

$\dfrac{\mathrm{d}^2h}{\mathrm{d}t^2} = -10 < 0$, a maximum.

$h = 40 - 20 = 20$ m.

$V = x(12-2x)^2 = x\left(144 - 48x + 4x^2\right) = 144x - 48x^2 + 4x^3$

$\dfrac{\mathrm{d}V}{\mathrm{d}x} = 144 - 96x + 12x^2 = 12\left(x^2 - 8x + 12\right) = 12(x-2)(x-6)$

$x = 2$ or $x = 6$. Reject $x = 6$, since the base would be zero.

$\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2} = -96 + 24x$; at $x = 2$ this is $-48 < 0$, a maximum.

$x = 2$ cm, giving $V = 2 \times 64 = 128$ cm³.

$2x + y = 100$, so $y = 100 - 2x$ and $A = 100x - 2x^2$.

$\dfrac{\mathrm{d}A}{\mathrm{d}x} = 100 - 4x = 0$, so $x = 25$.

$y = 50$, and $A = 25 \times 50 = 1250$ m².

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12 - 3x^2 = 0$, so $x^2 = 4$ and $x = 2$ (in range).

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -6x = -12 < 0$, a maximum.

$y = 24 - 8 = 16$.

Endpoints: $y(0) = 0$ and $y(3) = 36 - 27 = 9$, both less than $16$.

Greatest value $16$, at $x = 2$.

$\dfrac{\mathrm{d}P}{\mathrm{d}n} = 60 - 2n = 0$ gives $n = 30$ — which is outside the range.

So the maximum must be at an endpoint. Since $P$ is increasing for $n < 30$, the largest value in range is at $n = 25$:

$P = 1500 - 625 - 200 = 675$

Maximum profit £$675$ at $n = 25$.

The width is $2t$ and the height is $9 - t^2$.

$A = 2t\left(9 - t^2\right) = 18t - 2t^3$

$\dfrac{\mathrm{d}A}{\mathrm{d}t} = 18 - 6t^2 = 0$, so $t^2 = 3$ and $t = \sqrt3$ (taking $t > 0$).

$\dfrac{\mathrm{d}^2A}{\mathrm{d}t^2} = -12t < 0$ for $t > 0$, a maximum.

$A = 18\sqrt3 - 2\left(3\sqrt3\right) = 12\sqrt3 = 20.8$ (3 s.f.)

(a) A closed cylinder has two circular ends and a curved side:

$S = 2\pi r^2 + 2\pi r h = 600\pi$

Divide through by $2\pi$: $r^2 + rh = 300$

$rh = 300 - r^2$, so $h = \dfrac{300 - r^2}{r}$  as required.

(b) $V = \pi r^2 h = \pi r^2 \times \dfrac{300-r^2}{r}$

$= \pi r\left(300 - r^2\right) = \pi\left(300r - r^3\right)$  as required.

Note this is now a polynomial in $r$, so the standard differentiation rule applies.

(c) $\dfrac{\mathrm{d}V}{\mathrm{d}r} = \pi\left(300 - 3r^2\right) = 0$

$r^2 = 100$, so $r = 10$ (rejecting $r = -10$, as a radius cannot be negative).

$\dfrac{\mathrm{d}^2V}{\mathrm{d}r^2} = \pi(-6r) = -60\pi < 0$ at $r = 10$, so this is a maximum.

(d) $V = \pi\left(3000 - 1000\right) = 2000\pi = 6280$ cm³ (3 s.f.)

$h = \dfrac{300 - 100}{10} = 20$ cm.

Comment: the height $20$ cm is exactly twice the radius $10$ cm, so $h = 2r$ — the tin's height equals its diameter.

This is the classic result: for a fixed surface area, the cylinder of greatest volume is as tall as it is wide, the shape closest to a sphere that a cylinder can manage. It is why tins of food are roughly this proportion, though in practice manufacturing and stacking costs pull the shape a little off the mathematical optimum.