Question

the given functions provide the connection between possible a rectangle for a given side length, x, and a set amount of peri different amounts of area you can close in with a given amou 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(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.
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Explanation:

Step1: Calculate \( A(20) \)

Substitute \( x = 20 \) into \( A(x)=x(75 - x) \).

$$
LATEXBLOCK0
$$

Step2: Calculate \( A(35) \)

Substitute \( x = 35 \) into \( A(x)=x(75 - x) \).

$$
LATEXBLOCK1
$$

Step3: Calculate \( A(40) \)

Substitute \( x = 40 \) into \( A(x)=x(75 - x) \).

$$
LATEXBLOCK2
$$

Step4: Solve \( A(x)=0 \)

Set \( x(75 - x)=0 \).
Using the zero - product property \( ab = 0\) implies \(a = 0\) or \(b = 0\).
So \(x=0\) or \(75 - x=0\), which gives \(x = 0\) or \(x = 75\).

Step5: Find the maximum of \( A(x)=x(75 - x)=-x^{2}+75x \)

The function \(y = ax^{2}+bx + c\) (here \(a=-1\), \(b = 75\), \(c = 0\)) has its vertex at \(x=-\frac{b}{2a}\).

$$ x=-\frac{75}{2\times(-1)}=\frac{75}{2}=37.5 $$

Since \(a=-1<0\), the parabola opens downwards and the vertex is the maximum point.

Answer:

a. \(A(20)=1100\)
b. \(A(35)=1400\)
c. \(A(40)=1400\)
d. \(x = 0\) or \(x = 75\)
e. \(A(x)\) is at its maximum when \(x = 37.5\) because for the quadratic function \(A(x)=-x^{2}+75x\) (where \(a=-1\), \(b = 75\)), the \(x\) - coordinate of the vertex (which is the maximum since \(a<0\)) is given by \(x=-\frac{b}{2a}=-\frac{75}{2\times(-1)} = 37.5\)