Chapter 4, Problem 47RE

To determine

To find : The dimension.

Answer to Problem 47RE

The answer:

  $4\text{ft}\times 4\text{ft}\times 2\text{ft}$

Explanation of Solution

Given information:

An open top rectangular stainless steel have to a square base and volume $32{\text{ft}}^3$ of sheet steel.

Calculation:

Examining the exercise reveals that this topic falls under the heading of chapter "Modelling and Optimization," thus we will apply the techniques and concepts discussed there.

To use the following notation assuming the stated circumstances.

  $x=$ Side of the base

  $h=$ Height of our vat

  $S=$ Surface of our vat

  $V=$ Volume of our vat

Formulas are derived from the exercise and notation.

  $\begin{array}{c}S=4xh+x^2\\ V=h{x}^2\\ =32\end{array}$

First simplify it by expressing the height from the volume because we know that for the weight to be at its minimum, our surface of the vat must also be at its minimum.

  $\begin{array}{c}h{x}^2=32\\ \to h=\frac{32}{x^2}\\ S=4\cdot \frac{32}{x^2}\cdot x+x^2\\ =\frac{128}{x}+x^2\\ S(x)=\frac{128}{x}+x^2\\ \to S^{\prime }(x)=-\frac{128}{x^2}+2x\\ =0\\ -\frac{128}{x^2}+2x=0\\ -128+2x^3=0\\ x^3=64\\ \to x=4\end{array}$

Now substituting our $x$ back into our formula for the volume get $h$ :

  $\begin{array}{c}h=\frac{32}{x^2}\\ =\frac{32}{16}\\ =2\end{array}$

Therefore, the required dimension: $4\text{ft}\times 4\text{ft}\times 2\text{ft}$