Question

Architecture A special window has the shape of a rectangle surmounted by an equilateral triangle. See the figure. If the perimeter of the window is $16$feet, what dimensions will admit the most light?

[Hint: Area of an equilateral triangle $\left(\frac{\sqrt{3}}{4}\right)x^2,$ where $x$ is the length of a side of the triangle.]

Short answer

The dimension will admit the most at$14.95f{t}^2.$

Step-by-step solution

Step 1. Given information

It is given that a special window has the shape of a rectangle surrounded by an equilateral triangle if the perimeter of the window is $16ft.$We need to determine the dimension that will admit the most.

Step 2. Solving

The perimeter of the window is calculated by,

$P=3x+2a.$

Put $P=16$in $P=3x+2a.$

$16=3x+2a.$

$2a=16-3x.$

$a=8-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2x$}\right..$

The area of the window $\left(A\right)$is determined by the sum of the area of an equilateral triangle $\left(A_t\right)$and the area of a rectangle.

$\left(A_t\right)=\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4x^2$}\right..$

$\left(A_r\right)=a·x.$

$A=\left(A_t\right)+\left(A_r\right).$

localid="1647342292315" $\raisebox{1ex}{$=\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4x^2+ax.$}\right.$

$=\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4x^2+\left(8-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2x$}\right.}\right)x.$}\right.$

$=\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4x^2+8x-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2x^2$}\right.}.$}\right.$

$A=-1.07x^2+8x.$

The vertex of the function is determined by,

$x=-\frac{8}{2·\left(-107\right)}.$

$=3.74ft.$

For the value of $x=3.74ft,$ the maximum window area is,

$A=-1.07{\left(3.74\right)}^2+8\left(37.74\right).$$A=-1.07{\left(3.74\right)}^2+8\left(37.74\right).$

$=14.95f{t}^2.$