199k views
3 votes
The junior class leadership students are making a homecoming poster. A diagram of the poster is shown below. The width is 5x and the height is 4x, where x is in inches. 2 inches of the material on each side is black.

a) Define a function a(x) for the area of the white space.
b) What is the maximum area that the white part of the banner can be if the maximum width of the white space is 10 inches? Show your work.

1 Answer

5 votes

Answer:

a) To define the function a(x) for the area of the white space, we need to calculate the total area of the poster and subtract the area of the black sides.

The total area of the poster is given by the product of its width and height:

Total Area = Width * Height = (5x) * (4x) = 20x^2 square inches

The black sides have a width of 2 inches on each side, so the width of the white space is reduced by 4 inches:

Width of White Space = Width - Black Sides = (5x) - 4 = 5x - 4 inches

Similarly, since there are black sides on both ends, the height of the white space is reduced by 4 inches as well:

Height of White Space = Height - Black Sides = (4x) - 4 = 4x - 4 inches

Now, we can calculate the area of the white space by multiplying its width and height:

Area of White Space = Width of White Space * Height of White Space

= (5x - 4) * (4x - 4)

= 20x^2 - 24x + 16 square inches

Therefore, the function a(x) for the area of the white space is:

a(x) = 20x^2 - 24x + 16

b) To find the maximum area that the white part of the banner can be if the maximum width of the white space is 10 inches, we need to maximize the function a(x).

To find the maximum or minimum point of a quadratic function, we can use calculus. Taking the derivative of a(x) with respect to x and setting it equal to zero will give us critical points. We can then determine if these critical points correspond to a maximum or minimum by analyzing their concavity.

Taking the derivative of a(x):

a'(x) = d/dx (20x^2 - 24x + 16)

= 40x - 24

Setting a'(x) equal to zero and solving for x:

40x - 24 = 0

40x = 24

x = 24/40

x = 0.6 inches

To determine if this critical point corresponds to a maximum or minimum, we can analyze the concavity of the function. Taking the second derivative of a(x) will help us with this:

a''(x) = d^2/dx^2 (20x^2 - 24x + 16)

= d/dx (40x - 24)

= 40

Since the second derivative is positive (40), it indicates that the function is concave up, which means the critical point corresponds to a minimum.

Therefore, the maximum area that the white part of the banner can be is obtained when x = 0.6 inches. Plugging this value into the function a(x):

a(0.6) = 20(0.6)^2 - 24(0.6) + 16

= 7.2 square inches

So, the maximum area of the white space is 7.2 square inches when the maximum width of the white space is 10 inches.

Explanation:

User LietKynes
by
7.7k points