29 - 31 Solved problems in maxima and minima

$a = 4x + y$

$y = a - 4x$
 

Volume
$V = x^2 y$

$V = x^2 \, (a - 4x)$

$V = ax^2 - 4x^3$
 

$dV/dx = 2ax - 12x^2 = 0$

$2x \, (a - 6x) = 0$
 

For 2x = 0; x = 0 (meaningless)

For a - 6x = 0; x = 1/6 a

Use x = 1/6 a
 

$y = a - 4(\frac{1}{6}a)$

$y = \frac{1}{3}a$
 

$V_{max} = (\frac{1}{3}a)^2 (\frac{1}{3}a)$

$V_{max} = \frac{1}{108} a^3 \, \text{ in}^3$           answer

From the figure:
$D^2 = d^2 + h^2$

$0 = 2d + 2h \dfrac{dh}{dd}$

$\dfrac{dh}{dd} = -\dfrac{d}{h}$
 

Volume of cylinder:
$V = \frac{1}{4}\pi d^2 h$

$\dfrac{dV}{dh} = \dfrac{\pi}{4} \left[ d^2 \dfrac{dh}{dd} + 2dh \right] = 0$

$d \dfrac{dh}{dd} + 2h = 0$

$d\left( -\dfrac{d}{h} \right) + 2h = 0$

$2h = \dfrac{d^2}{h}$

$d^2 = 2h^2$

$d = \sqrt{2} \, h$
 

$\text{diameter } \, = \sqrt{2} \, \times \, \text{ height }$           answer

Convex surface area (shaded area):
$A = \pi dh$

$\dfrac{dA}{dd} = \pi \left( d \dfrac{dh}{dd} + h \right) = 0$

$d \dfrac{dh}{dd} + h = 0$
 

From Solution to Problem 30 above, dh/dd = -d/h
$d \left( -\dfrac{dh}{dd} \right) + h = 0$

$h = \dfrac{d^2}{h}$

$h^2 = d^2$

$d = h$

$\text{diameter } = \text{ height}$           answer