Rectangle with given perimeter: maximum area

Text

A rectangle has perimeter $24$; its base is $x$.

• (a) show that the height $h$ depends on $x$ like a line and write it; what is $h$ for $x=1$ and $x=3$?
• (b) write the area as a function of $x$; what is it for $x=1$ and $x=3$?
• (c) show that the area is a downward-opening parabola, find its vertex, and deduce the maximum area and the corresponding $x$.

Solution

(a) Height. The perimeter is $2(x+h)=24$, so $x+h=12$ and $h=12-x$ This is a line (a first-degree function of $x$). In particular $h(1)=11$ and $h(3)=9$.

(b) Area. $A(x)=x\cdot h=x(12-x)=-x^2+12x$. In particular $A(1)=1\cdot11=11$ and $A(3)=3\cdot9=27$.

(c) Maximum area. $A(x)=-x^2+12x$ has $a=-1<0$: it is a downward-opening parabola. The vertex is $x_V=-\frac{b}{2a}=-\frac{12}{2(-1)}=6, \qquad A(6)=-36+72=36$ The maximum area is $36$, obtained at $x=6$ (the rectangle is then a $6\times6$ square). $\boxed{h=12-x\quad A=-x^2+12x\quad x_{\max}=6,\ A_{\max}=36}$