Question

Find the length and width of a rectangle that has the given perimeter and a maximum area. Perimeter: 100 meters

Short answer

The dimensions that give the maximum area for a rectangle with a perimeter of 100 meters are a length of 25 meters and a width of 25 meters. This makes the rectangle a square, which is expected since among all rectangles with a given perimeter, a square has the maximum area.

Step-by-step solution

Write the formulas

Write down the formulas for the perimeter and area of a rectangle. The perimeter \(P\) is \(2l + 2w\) and the area \(A\) is \(lw\), where \(l\) and \(w\) are the rectangle's length and width respectively.

Express the width in terms of the perimeter and length

We are given that the perimeter \(P\) is 100 meters, so we have \(2l + 2w = 100\). We can solve this equation for \(w\) to express it as a function of \(l\): \(w = (100 - 2l)/2 = 50 - l\).

Express the area in terms of the length

Substitute \(w\) into the area formula to write \(A\) as a function of \(l\): \(A = l(50 - l)\).

Find the maximum area

To find the maximum area, we need to maximize the function \(A(l) = l(50 - l)\). This is a parabola that opens downwards, and its maximum value occurs at its vertex. The x-coordinate of the vertex of a parabola given by \(y = a(x - h)^2 + k\) is \(h\), so in this case, \(l = 50/2 = 25\).

Calculate width using the optimal length

Substitute \(l = 25\) into the equation for \(w\) that we found in Step 2 to find the width that gives the maximum area: \(w = 50 - 25 = 25\).