Problem 6 - Calculus Early Transcendentals

Question

Of all rectangles with a fixed perimeter of \(P,\) which one has the maximum area? (Give the dimensions in terms of \(P .\) )

Step-by-Step Solution

Answer

Answer: The rectangle with maximum area is a square, with each side length equal to \(\frac{P}{4}\), where \(P\) is the fixed perimeter.

Step 1: Write down the problem constraints.

The problem states that the perimeter of the rectangle is fixed at \(P\). We know that the perimeter of a rectangle is given by: \(P = 2(L + W)\) We want to find the dimensions, i.e., the length \(L\) and the width \(W\), that maximize the area of the rectangle, which is given by: \(A = L \cdot W\)

Step 2: Express the width W in terms of the length L.

In order to find the relationship between the length \(L\) and the width \(W\), we can use the formula for the perimeter. First, solve the perimeter formula for width \(W\): \(W = \frac{P - 2L}{2}\)

Step 3: Substitute W in terms of L into the area formula.

Now, we can replace the width \(W\) in the area formula with the expression we found in the previous step: \(A = L \cdot \frac{P - 2L}{2}\)

Step 4: Simplify the area expression.

First, notice that we can cancel out the factor \(2\): \(A = L \cdot \frac{P - 2L}{2} \Rightarrow A = L \cdot \frac{P - 2L}{1}\) Now we get the simplified expression for the area: \(A = L(P - 2L)\)

Step 5: Observe how the area behaves with varying L.

The expression \(A = L(P - 2L)\) is a quadratic that is concave down (since the coefficient of the square term is negative), and it attains a maximum value when: \(L = \frac{P - 2L}{2} \Rightarrow L = \frac{P}{4}\)

Step 6: Determine the value of W corresponding to the optimal L.

We can now use the value of \(L = \frac{P}{4}\) to determine the optimal value for the width \(W\) by substituting into the formula for \(W\) in terms of \(L\): \(W = \frac{P - 2L}{2} = \frac{P - 2(\frac{P}{4})}{2} = \frac{P}{4}\)

Step 7: Conclusion

From our findings, the dimensions that provide the maximum area for a rectangle with a fixed perimeter \(P\) are \(L = \frac{P}{4}\) and \(W = \frac{P}{4}\). This means that the rectangle with maximum area is actually a square, with each side length equal to \(\frac{P}{4}\).