Question

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topic: symmetry and distance
the given functions provide the connection between possible areas, ( a(x) ), that can be created by a rectangle for a given side length, ( x ), and a set amount of perimeter. you could think of it as the different amounts of area you can close in with a given amount of fencing as long as you always create a rectangular enclosure.

1. ( a(x)=x(10 - x) )

find the following:
a. ( a(3)= )
b. ( a(4)= )
c. ( a(6)= )
d. ( a(x)=0 )
e. when is ( a(x) ) at its maximum? explain or show how you know.

1. ( a(x)=x(50 - x) )

find the following:
a. ( a(10)= )
b. ( a(20)= )
c. ( a(30)= )
d. ( a(x)=0 )
e. when is ( a(x) ) at its maximum? explain or show how you know.

1. ( a(x)=x(75 - x) )

find the following:
a. ( a(20)= )
b. ( a(35)= )
c. ( a(40)= )
d. ( a(x)=0 )
e. when is ( a(x) ) at its maximum? explain or show how you know.

1. ( a(x)=x(48 - x) )

find the following:
a. ( a(10)= )
b. ( a(20)= )
c. ( a(28)= )
d. ( a(x)=0 )
e. when is ( a(x) ) at its maximum? explain or show how you know.
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Explanation:

Step1: Expand the function

The function \(A(x)=x(48 - x)=48x-x^{2}\). This is a quadratic function in the form \(y = ax^{2}+bx + c\), where \(a=- 1\), \(b = 48\), \(c = 0\).

Step2: Use the formula for the vertex of a quadratic function

For a quadratic function \(y=ax^{2}+bx + c\), the \(x\) - coordinate of the vertex (where the function reaches its maximum or minimum) is given by \(x=-\frac{b}{2a}\).
Substitute \(a=-1\) and \(b = 48\) into the formula:

$$x=-\frac{48}{2\times(-1)}=\frac{-48}{-2}=24$$

Answer:

The function \(A(x)=x(48 - x)\) reaches its maximum when \(x = 24\).