Question

The Shortest Fence \(\mathrm{A} 216-\mathrm{m}^{2}\) rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?

Short answer

The dimensions for the rectangle that will require the smallest total length of fence are 18m by 12m and the total fence required is 72m.

Step-by-step solution

Set up the function

Assign variables to the length and width of the rectangle, say \(x\) and \(y\) respectively. The area of the rectangle is given by the product of length times the width, \(x*y = 216\). Given that the rectangle is divided into two equal parts by a fence parallel to one side (assuming \(y\) side), the total length of fence is \(F = 2x + 3y\). We need F to be a function of a single variable, so we solve the area equation for \( y = 216 / x \) and substitute this into the function F.

Find the derivative of the function

Now our function \(F(x) = 2x + 3(216 / x)\). To find out where this function is minimum, we need to find its critical points. These will happen where its derivative is zero or undefined. Derivate of the function \(F'(x) = 2 - 648/x^2 \).

Find the critical points

Setting the derivative equal to zero and solving for \(x\), we find that \(0 = 2 - 648/x^2 \). Solving this equation leads to \(x = \sqrt{324}\), which is \(x = 18\). Checking the end points, as x approaches 0, F goes to infinity. As x approaches infinity, F also goes to infinity. So, x = 18 is the only minima.

Substitute the value of \(x\) to find \(y\) and find the total length of fence

Substitute \(x = 18\) in width equation to find \(y = 216 / x = 12\). So, the dimensions of the rectangle that would require the smallest total length of fence are 18m by 12m. Total length of fence needed is \(F(18) = 2*18 + 3*12 = 72m\).