A homebuyer is building a home and has sat down with an architect. The buyer tells the architect the perimeter of the house must be 200
feet. The dimensions of the house that would give the buyer the maximum area would be which of the following?
A 10 x 90 feet
B 20 x 80 feet
C 30 x 70 feet
40 x 60 feet
E 50 x 50 feet

[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

[tex]f^{\prime}(x) = -2\, x + 100[/tex] and
[tex]f^{\prime\prime}(x) = -2[/tex].

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.