a rectangular box with two sqaure opposite ends is to hold 8000 cubic inches. find the dimensions of the cheapest box if tge recangular sides cost 15 times more as much per square inche as the top, bottom, and square ends

the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in

Step-by-step explanation:

since the box has a volume V

V = x*y*z = b=8000 in³

since y=z (square face)

V = x*y² = b=8000 in³

and the cost function is

cost = cost of the square faces * area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides

C = a * 2*y² + a * 2*x*y + 15*a * 2*x*y = 2*a * y² + 32*a*x*y

to find the optimum we can use Lagrange multipliers, then we have 3 simultaneous equations:

x*y*z = b

Cx - λ*Vx = 0 → 32*a*y - λ*y² = 0 → y * (32*a-λ*y) = 0 → y=32*a/λ

Cy - λ*Vy = 0 → (4*a*y + 32*a*x) - λ*2*x*y = 0

4*a*32/λ + 32*a*x - λ*2*x*32*a/λ = 0

128*a² / λ + 32*a*x - 64*a*x = 0

32*a*x = 128*a² / λ

x = 4*a/λ

x*y² = b

4*a/λ * (32*a/λ) ² = b

(a/λ) ³ * 4096 = 8000 m³

(a/λ) = ∛ (8000 m³/4096) = 5/4 in

then

x = 4*a/λ = 4*5/4 in = 5 in

y=32*a/λ = 32*5/4 in = 40 in

then the box has dimensions 5 in x 40 in x 40 in