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A 20,000 m^3 rectangular cistern is to be made from reinforced concrete such that the interior length will be twice the height. If the cost is $40/m^2 for the base, $100/m^2 for the side walls, and $200/m^2 for the roof, find the interior dimensions (to one decimal place) that will keep the cost to a minimum. To protect the water table, the building code specifies that no excavation can be more than 22m deep. It also specifies that all cisterns must be at least 1m deep

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The volume of 20,000 m³, and the cost of $40/m², $100/m², and

$200/m², give the dimensions that give minimum cost as follows;

  • The height of the cistern is approximately 20.4 meters
  • The length is approximately 40.8 meters
  • The width is approximately 24.029 meters

How can the dimensions of the cistern be found?

The dimensions of the cistern = L × W × H

L = 2·H

L × W × H = 20000

2·H² × W = 20000

[tex]W = \mathbf{ \dfrac{20000}{2 \cdot H^2}}[/tex]

Therefore;

A = 40·L·W + 200·L·W + 100·2·L·H + 100·2·W·H

A = 400·H² + 680·W·H

Given that the leading coefficient is positive, the function has a minimum point.

[tex]A = \dfrac{400 \cdot H^4 + 6800000 \cdot H}{H^2}[/tex]

[tex]\dfrac{d}{dH} A = \mathbf{\dfrac{d}{dH} \left( \dfrac{400 \cdot H^4 + 6800000 \cdot H}{H^2} \right)} = 0[/tex]

Which gives;

800·H⁵ - 6800000·H² = 0

800·H⁵ = 6800000·H²

H³ = 6800000 ÷ 800 = 8500

H ≈ 20.4 m

Which gives;

L ≈ 2 × 20.4 m =  40.8 m

[tex]W = \dfrac{20000}{2 \times 20.4^2} \approx \mathbf{24.029}[/tex]

The interior dimensions of the cistern are;

  • The height, H ≈ 20.4 m.
  • The length, L ≈ 40.8 m
  • Width of the cistern, W ≈ 24.029 m

Learn more about minimum value of a function here:

https://brainly.com/question/73849

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