Answer:
[tex]50\; {\rm ft} \times 50\; {\rm ft}[/tex] would maximize the area for a rectangle with the given circumference of [tex]200\; {\rm ft}[/tex]. (Note, that a circle of the same circumference would have an even larger area.)
Step-by-step explanation:
Assume that the base of the house is a rectangle. Let the length of the two sides be [tex]x\; {\rm ft}[/tex] and [tex]y\; {\rm ft}[/tex], respectively. The goal is to find the [tex]x[/tex] and [tex]y[/tex] that:
[tex]\begin{aligned} \text{maximize} \quad & x\, y \\ \text{subject to} \quad & 2\, (x + y) = 200 \\ & x > 0 \\ & y > 0 \end{aligned}[/tex].
Using the equality constraint [tex]2\, (x + y) = 200[/tex] (or [tex]x + y = 100[/tex]), the variable [tex]y[/tex] could be replaced with [tex](100 - x)[/tex] to obtain an equivalent problem of only one variable:
[tex]\begin{aligned} \text{maximize} \quad & x\, (100 - x) \\ \text{subject to} \quad & x > 0 \\ & (100 - x) > 0 \end{aligned}[/tex].
Simplify to obtain:
[tex]\begin{aligned} \text{maximize} \quad & -x^{2} + 100\, x \\ \text{subject to} \quad & x > 0 \\ & x < 100 \end{aligned}[/tex].
The objective function of this problem is [tex]f(x) = -x^{2} + 100\, x[/tex]. Derivatives of this function include
Since [tex]f^{\prime\prime}(x)[/tex] is constantly less than [tex]0[/tex], [tex]f(x)[/tex] is concave and would be maximized when [tex]f^{\prime}(x) = 0[/tex].
Setting [tex]f^{\prime}(x) = -2\, x + 100[/tex] to [tex]0[/tex] and solving for [tex]x[/tex] gives:
[tex]-2\, x + 100 = 0[/tex].
[tex]x = 50[/tex].
Notice that [tex]x = 50[/tex] satisfies both constraints: [tex]x > 0[/tex] and [tex]x < 100[/tex]. Therefore, [tex]x = 50[/tex] is indeed the solution that maximizes the area [tex]f(x) = -x^{2} + 100\, x[/tex] while at the same time meeting the requirements.
With the length of one side being [tex]x = 50[/tex] ([tex]50\; {\rm ft}[/tex],) the length of the other side would be [tex]100 - x = 50[/tex] ([tex]50\; {\rm ft}\![/tex].) Hence, a rectangular house of dimensions [tex]50\; {\rm ft} \times 50\; {\rm ft}[/tex] would maximize the area under the given requirements.
