a rectangular tank that is 2916 ft³ with a square base and open top is to be constructed of sheet steel of a…

Explanation:

Step1: Define variables

Let the side length of the square base be (x) (in feet) and the height of the tank be (y) (in feet). The volume (V=x^{2}y), and given (V = 2916), so (y=\frac{2916}{x^{2}}). The surface - area (S) (since the top is open) is (S=x^{2}+4xy).

Step2: Substitute (y) into the surface - area formula

Substitute (y=\frac{2916}{x^{2}}) into (S): (S=x^{2}+4x\cdot\frac{2916}{x^{2}}=x^{2}+\frac{11664}{x}), where (x>0).

Step3: Find the derivative of (S) with respect to (x)

Using the power rule, if (S(x)=x^{2}+11664x^{- 1}), then (S^\prime(x)=2x - 11664x^{-2}=\frac{2x^{3}-11664}{x^{2}}).

Step4: Find the critical points

Set (S^\prime(x) = 0), so (2x^{3}-11664 = 0). (x^{3}=\frac{11664}{2}=5832), then (x=\sqrt[3]{5832}=18).

Step5: Use the second - derivative test

Find the second - derivative (S^{\prime\prime}(x)=2 + 23328x^{-3}). When (x = 18), (S^{\prime\prime}(18)=2+\frac{23328}{18^{3}}=2 + 4=6>0). So (S(x)) has a minimum at (x = 18).

Step6: Find the value of (y)

Substitute (x = 18) into (y=\frac{2916}{x^{2}}), (y=\frac{2916}{18^{2}}=\frac{2916}{324}=9).

Answer:

The dimensions of the tank are (18) ft (side of the square base) and (9) ft (height).