1. Sam needs to construct two adjacent pens against a wall as show below. The pens must enclose a total of 300 square feet. What is the minimum length of fence required to enclose the pens? (No fence is required against the wall.) Use calculus - either the first or second derivative test - to verify your result.

Step 1 ：Let's denote the length of the fence perpendicular to the wall as \(x\), and the length of the fence parallel to the wall as \(y\). Then the total length of the fence is \(2x + y\), and the total area enclosed is \(xy = 300\).

Step 2 ：We can express \(y\) in terms of \(x\) using the area constraint: \(y = 300/x\).

Step 3 ：Then we can express the total length of the fence in terms of \(x\) only: \(L(x) = 2x + 300/x\).

Step 4 ：To find the minimum length, we need to find the derivative of \(L(x)\), set it equal to zero, and solve for \(x\). The derivative of \(L(x)\) is \(L'(x) = 2 - 300/x^2\).

Step 5 ：The critical points are \(-5\sqrt{6}\) and \(5\sqrt{6}\). However, since \(x\) represents a length, it cannot be negative. So the only valid critical point is \(x = 5\sqrt{6}\).

Step 6 ：We can substitute this value back into the equation for \(L\) to find the minimum length of the fence. The minimum length of fence required to enclose the pens is \(\boxed{20\sqrt{6}}\) feet.