Chapter 3.4, Problem 18AYU

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 $=\left(\frac{\surd 3}{4}\right)x^2$, where $x$ is the length of a side of the triangle.]

To determine

To calculate: The dimension of the window which admits the maximum light.

Answer to Problem 18AYU

Solution:

The dimension of the window is $3.74$ feet by $2.39$ feet.

Explanation of Solution

Given:

The perimeter of the window is 16 feet.

The shape of the window is

Formula used:

The area of the equilateral triangle is $\frac{\sqrt{3}}{4}{x}^2$.

The area of the window is $A=xy+\frac{\sqrt{3}}{4}{x}^2$.

Perimeter of the window is $P=3x+2y$.

Calculation:

Here, we have to find the dimension of the window with maximum area.

The perimeter of the window is 16 feet.

Thus, we have

$16=3x+2y$

$\Rightarrow y=\frac{16-3x}{2}$

$\Rightarrow y=8-1.5x$

Thus, we get the area of the window as

$A=x(8-1.5x)+\frac{\sqrt{3}}{4}{x}^2$

$=8x-1.5x^2+\frac{\sqrt{3}}{4}{x}^2$

$=8x-1.067x^2$

Thus, we can see that the area is a quadratic equation with $a=-1.067,b=8,c=0$.

Since, $a$ is negative, the maximum value is found at the vertex.

Therefore, the point of the maximum area is

$x=-\frac{b}{2a}=-\frac{8}{2(-1.067)}=3.74$

Now, we need to find the value of $y$.

Thus, we have

$y=8-1.5(3.74)$

$=2.39$

Therefore, the dimension of the window is feet by feet.