Question

a rectangular tank with a square base, an open top, and a volume of 32 ft³ is to be constructed of sheet steel. find the dimensions of the tank that has the minimum surface area.
the dimensions of the tank with minimum surface area are □ ft.
(simplify your answer. use a comma to separate answers.)

Explanation:

Step1: Set up the volume and surface area equations

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 since \(V = 32\), we have \(y=\frac{32}{x^{2}}\).
The surface area \(S=x^{2}+4xy\) (because the tank has an open - top, so one base area \(x^{2}\) and four lateral faces each of area \(xy\)).

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

Substitute \(y=\frac{32}{x^{2}}\) into \(S\):
\(S=x^{2}+4x\cdot\frac{32}{x^{2}}=x^{2}+\frac{128}{x}\), where \(x>0\).

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

Using the power rule, if \(S(x)=x^{2}+\frac{128}{x}=x^{2}+128x^{-1}\), then \(S^\prime(x)=2x - 128x^{-2}=\frac{2x^{3}-128}{x^{2}}\).

Step4: Set the derivative equal to zero and solve for \(x\)

Set \(S^\prime(x) = 0\), so \(\frac{2x^{3}-128}{x^{2}}=0\). Since \(x^{2}>0\) for \(x>0\), we solve \(2x^{3}-128 = 0\).
\(2x^{3}=128\), \(x^{3}=64\), \(x = 4\).

Step5: Find the value of \(y\)

Substitute \(x = 4\) into \(y=\frac{32}{x^{2}}\).
\(y=\frac{32}{4^{2}}=\frac{32}{16}=2\).

Answer:

\(4,2\)