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.
2 ( 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.
3 ( 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.
4 ( 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: Calculate \(A(10)\) for \(A(x)=x(50 - x)\)

Substitute \(x = 10\) into the function:
\(A(10)=10\times(50 - 10)=10\times40 = 400\)

Step2: Calculate \(A(20)\) for \(A(x)=x(50 - x)\)

Substitute \(x = 20\) into the function:
\(A(20)=20\times(50 - 20)=20\times30=600\)

Step3: Calculate \(A(30)\) for \(A(x)=x(50 - x)\)

Substitute \(x = 30\) into the function:
\(A(30)=30\times(50 - 30)=30\times20 = 600\)

Step4: Solve \(A(x)=0\) for \(A(x)=x(50 - x)\)

Set \(x(50 - x)=0\). By the zero - product property, if \(ab = 0\), then \(a = 0\) or \(b = 0\).
So \(x=0\) or \(50 - x=0\), which gives \(x = 0\) or \(x = 50\)

Step5: Find the maximum of \(A(x)=x(50 - x)\)

The function \(A(x)=x(50 - x)=-x^{2}+50x\) is a quadratic function in the form \(y = ax^{2}+bx + c\) (\(a=-1\), \(b = 50\), \(c = 0\)).
The \(x\) - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\).
Substitute \(a=-1\) and \(b = 50\) into \(x=-\frac{b}{2a}\):
\(x=-\frac{50}{2\times(-1)}=\frac{-50}{-2}=25\)

Answer:

a. \(A(10)=400\)
b. \(A(20)=600\)
c. \(A(30)=600\)
d. \(x = 0\) or \(x = 50\)
e. \(A(x)\) is at its maximum when \(x = 25\) because for a quadratic function \(y=-x^{2}+50x\) (where \(a=-1\), \(b = 50\)), the \(x\) - coordinate of the vertex \(x=-\frac{b}{2a}=25\) and since \(a=-1<0\), the parabola opens downwards.