Please help applications of differentiation .. I don’t get it 🥲🥲

Question

A manufacturer of open steel boxes has to make a box with a square base and a capacity of 1 kL. The steel costs $24 per square metre.

a)if the base measures x m by x m and the height is y m, find y in terms of x.

b) Hence shows that the total cost of the steel is

C(x)=24x2+96/x dollars

c)Find C^l (x)

d) Hence find the dimensions which minimise the cost of the box, and the cost of the steel in this case.

Answer 1

This is a classic minimization problem

a)if the base measures x m by x m and the height is y m, find y in terms of x.

AreaBox = yx*4+2*x^2

VolumeBox = x*x*y = 1KL

y = 1K/x^2

AreaBox = 4*1k/x + 2*x^2

Rate = $24/m^2

C(x) = 24*4k/x + 24*2*x^2

dC/dx = -96k*x - 2*48/x = 0. Maxima

mux by - and div by 48

2k*x + 2/x = 0

Div by 2

kx + 1/x = 0

kx = -1/x

x^2 = -1/k

x = i * sqr(1/k)

x = I * 0.0316 m

Looks like you have an imaginary solution check the sign if I’ve got it right it’s imaginary else it’s a real number of 31.6 cm