A rectangular metal tank with an open top is to hold 171.5cubic feet of liquid. What are the dimensions of the tank that require the least material to build?

7 feet by 7 feet by 3.5 feet

Step-by-step explanation:

The volume of the tank which is open at the top = 171.5 ft³

Where:

I = length

w = width

h = height

V = Volume

The surface area of the tank because it is open at the top = lw + 2wh + 2lh Equation 1

Volume of the tank =lwh Equation 2

Hence,

h = V/lw Equation 3

Putting this into our Equation 1

lw + 2w(V/lw) + 2l(V/lw)

lw + 2V/l + 2V/w

Finding the derivative

dy/dl = w - 2V/l²

dy/dw = l - 2V/w²

Hence:

w - 2V/l² = 0

w = 2V/l² Equation 4

l - 2V/w² = 0

l = 2V/w² Equation 5

l = 2V/w² Equation 5

Note : w = 2V/l², so we substitute

l = 2V/(2V/l²)²

l = 2V ÷ (4V²/l⁴)²

I = 2V × l⁴/4V²

I = l⁴/2V

Cross Multiply

I × 2V = l⁴

Divide both sides by l

2V = l⁴/l

2V = l³

l³ = 2V

V = 171.5cm³

l = cube root(2 × 171.5)

l = cube root (343)

l = 7 feet

w = 2V/l² Equation 4

w = 2 × 171.5/7²

w = 343/49

w = 7 feet

h = V/lw Equation 3

h = 171.5/7 × 7

h = 171.5/49

h = 3.5 feet

Therefore, the dimensions of the tank that require the least material to build;

7 feet by 7 feet by 3.5 feet

Where:

7feet = Length

7 feet = Width

3.5 feet = Height