Answer and explanation:
Given : A rectangular frame of uniform depth for a shadow box is to be made from a 36-in. piece of wood.
To find :
a) Write a function to represent the display area in terms of x.
The perimeter of the box is 36 inches.
Let x and y be the length and width of the box.
So, ![2x+2y=36](https://tex.z-dn.net/?f=2x%2B2y%3D36)
![2y=36-2x](https://tex.z-dn.net/?f=2y%3D36-2x)
![y=\frac{36-2x}{2}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B36-2x%7D%7B2%7D)
....(1)
The area of the box is ![A=lb](https://tex.z-dn.net/?f=A%3Dlb)
![A=xy](https://tex.z-dn.net/?f=A%3Dxy)
Substitute y from (1),
![A=x(18-x)](https://tex.z-dn.net/?f=A%3Dx%2818-x%29)
![A=18x-x^2](https://tex.z-dn.net/?f=A%3D18x-x%5E2)
![A=-x^2+18x](https://tex.z-dn.net/?f=A%3D-x%5E2%2B18x)
The function in terms of x, ![A(x)=18x-x^2](https://tex.z-dn.net/?f=A%28x%29%3D18x-x%5E2)
b. What dimensions should be used to maximize the display area?
As A is a quadratic function with a negative leading coefficient.
The vertex is the maximum point on the function.
So, the x-coordinate of the vertex is the value of x that will maximize the area.
![x=\frac{-b}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%7D%7B2a%7D)
Here, a=-1 and b=18
![x=\frac{-18}{2(-1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-18%7D%7B2%28-1%29%7D)
![x=9](https://tex.z-dn.net/?f=x%3D9)
Substitute in (1),
![y=18-9](https://tex.z-dn.net/?f=y%3D18-9)
![y=9](https://tex.z-dn.net/?f=y%3D9)
The dimension 9 in. by 9 in. should be used maximize the display area.
c) What is the maximum area?
Substitute the value of x in the area,
![A(x)=-x^2+18x](https://tex.z-dn.net/?f=A%28x%29%3D-x%5E2%2B18x)
![A(9)=-9^2+18(9)](https://tex.z-dn.net/?f=A%289%29%3D-9%5E2%2B18%289%29)
![A(9)=-81+162](https://tex.z-dn.net/?f=A%289%29%3D-81%2B162)
![A(9)=81](https://tex.z-dn.net/?f=A%289%29%3D81)
The maximum area is 81 square inches.