56 - 57 Maxima and minima problems of square box and silo

Let
k = unit price of pine
mk = unit price of oak
C = total cost
 

Volume of the square box:
$V = x^2 \, y$

$\dfrac{dV}{dx} = x^2 \, y' + 2xy = 0$

$y' = -\dfrac{2y}{x}$
 

Total cost:
$C = kx^2 + kxy + mkx^2 + 3mkxy$

$C = k(m + 1)x^2 + k(3m + 1)xy$

$\dfrac{dC}{dx} = 2k(m + 1)x + k(3m + 1)(xy' + y) = 0$

$2(m + 1)x + (3m + 1) \, [ \, x(-2y/x) + y \, ] = 0$

$2(m + 1)x = (3m + 1)y$

$x = \dfrac{(3m + 1)y}{2(m + 1)}$           answer

Let
k = unit price
 

Total cost:
$C = [ \, \frac{1}{4}\pi D^2 + \pi D(H - \frac{1}{2}D) + \frac{1}{2}\pi D^2 \, ] \, k$

$C = [ \, \frac{1}{4}\pi D^2 + \pi DH \, ] \, k$

$\dfrac{dC}{dD} = \left[ \, \frac{1}{2}\pi D + \pi \left( D \dfrac{dH}{dD} + H \right) \, \right] \, k = 0$

$D + 2D \dfrac{dH}{dD} + 2H = 0$

$\dfrac{dH}{dD} = -\dfrac{2H + D}{2D}$
 

Volume of silo = volume of cylinder + volume of hemisphere:
$V = \frac{1}{4}\pi D^2(H - \frac{1}{2}D) + \frac{1}{12}\pi D^3$

$V = \frac{1}{4}\pi D^2 H - \frac{1}{24}\pi D^3$

$\dfrac{dV}{dD} = \frac{1}{4}\pi \left( D^2 \dfrac{dH}{dD} + 2DH \right) - \frac{1}{8}\pi D^2 = 0$

$2D \dfrac{dH}{dD} + 4H - D = 0$

$\dfrac{dH}{dD} = -\dfrac{D - 4H}{2D}$
 

$\dfrac{dH}{dD} = \dfrac{dH}{dD}$

$-\dfrac{2H + D}{2D} = -\dfrac{D - 4H}{2D}$

$-2H - D = D - 4H$

$H = D$
 

Total height = diameter           answer