Lois C. answered • 04/11/20
BA in secondary math ed with 20+ years of classroom experience
Since this is labeled as a calculus problem, we will use calculus techniques to solve it. The function to be maximized is the area of a rectangle, based on the formula A = l x w. We need to use the given perimeter to express both length and width in terms of the same variable, x. So let's let x = the width of the rectangle, and the length can be expressed as (40-2x)/2 or, in simplified form, 20-x. Using these two expressions in x to represent both dimensions of the rectangle, the area function becomes A(x) = x( 20-x ) or, once we simplify it, A(x) = -x2 + 20x.
To maximize this function, we take the derivative to find where A'(x) = 0. Since this is a quadratic function, this will occur at the vertex of the upside-down parabola. Taking the derivative, we have A'(x) = -2x + 20. Setting this equal to 0, we have x = 10. To confirm this, we can check that the derivative at x values to the left of 10 is positive ( indicating an increasing function to the left of the vertex) and that the derivative at x values to the right of 10 is negative ( indicating a decreasing function to the right of the vertex). Since this checks out, we know that the width = x = 10 and the length = 20-x = 10, so the maximum area will occur when the rectangle is actually a square.
