The base of an aquarium with given volume is made of slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials.

Base = Length = (2V/5)^⅓

Height = V^⅓/(2/5)^⅔

Step-by-step explanation:

Let the sides of the aquarium be l,b and h

Where l = length

b = base

h = height

Volume = lbh

V = lbh

The surface area of the box is

A = 2bh + 2lh + 2lb

We'll replace the 2lb with 5lb to get the cost of area because it's given the question that the slate (base) of the box cost 5 times as much per unit area as glass

C(l,b,h) = 2bh + 2lh + 5lb

Make h the subject of formula in (V = lbh)

h = V/lb

This will enable us to account for the lowest cost of materials by taking derivatives of the cost equation, and we need to solve for a l to put into the solution.

Substitute V/lb for h in the cost equation

C(l,b,V/lb) = 2b(V/lb) + 2l(V/lb) + 5lb Simplify

C(l,b,V/lb) = 2V/l + 2V/b + 5lb

Take the derivatives of the above with respect to l and b

Cl = -2V/l² + 5b

Cb = -2V/b² + 5l

Equate Cb to Cl (this implies that b = l)

Cb = Cl =>

-2V/l² + 5b = -2V/b² + 5l

So, we have

C(b,b) = -2V/b² + 5b = 0

-2V/b² + 5b = 0 Solve for b

-2V/b² = -5b Multiply through by b²

-2V = -5b³ Divide through by -5

2V/5 = b³ ---; Rearrange

b³ = 2V/5

b = (2V/5)^⅓

b = l

So, l = (2V/5)^⅓

h = V/lb (b = l)

h = V/l² Substitute (2V/5)^⅓ for l

h = V/((2V/5)^⅓)²

h = V/(2V/5)^⅔

h = V/((V^⅔)(2/5)^⅔)

h = V^⅓/(2/5)^⅔

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