Academic math help please

Question

Determine the dimensions of a square-based open topped prism with a volume of 24cm and a minimum surface area

Answer 1

This is a calculus problem. I'm assuming the prism looks like this:

The basic (primary) equation is the Surface Area equation.

SA = SA_Exterior + SA_Interior

SA = 2(SA_Exterior)

SA_Exterior = 4(Area of each lateral triangle) = 4(1/2*b*h_L)

SA = 2(4)(1/2*b*h_L) = 4bh_L

Secondary equation #1: (b/2)² + h_P² = h_L²

h_P = (h_L² + (b/2)²)^1/2

Secondary Equation #2:

V = 1/3b²h_P

Plugging Secondary Equation #1 into Secondary Equation #2:

V = 1/3b²(h_L² + (b/2)²)^1/2

But V = 24 so:

24 = 1/3b²(h_L² + (b/2)²)^1/2

72/b² = (h_L² + (b/2)²)^1/2

5184/b⁴ = h_L² + b²/4

h_L² = 5184/b⁴ - b²/4

h_L² = 20736/4b⁴ - b⁶/4b⁴

h_L² = (20736 - b⁶)/4b⁴

h_L = [(20736 - b⁶)/4b⁴]^1/2

h_L = (20736 - b⁶)^1/2/2b²

Plugging this into the Basic (primary) equation (SA = 4bh_L) we get:

SA(b) = 4b(20736 - b⁶)^1/2/2b²

SA(b) = 2(20736 - b⁶)^1/2/b

To minimize this function, take the derivative and set it equal to zero. To take the derivative, use the quotient rule (f'g - g'f)/g²

SA'(b) = (4b⁶ - 41472)/(b²(20736 + b⁶)^1/2

Set it equal to zero: 0 = (4b⁶ - 41472)/(b²(20736 + b⁶)^1/2

This can only be true if the numerator is equal to zero so:

0 = (4b⁶ - 41472)

b = 4.66963

Plugging that into the other equations:

h_L = 4.044

h_P = 3.30193