SOLUTION: Please help me solve this problem: Fred has 48m of fence. The area that can be enclosed by the fence is modeled by the function A(x)= -2x^2 +48x, where x is the width of the area i

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Question 882159: Please help me solve this problem: Fred has 48m of fence. The area that can be enclosed by the fence is modeled by the function A(x)= -2x^2 +48x, where x is the width of the area in meters and A(x)is the area in square meters. What is the maximum area that can be enclosed?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
: Fred has 48m of fence. The area that can be enclosed by the fence is modeled by the function A(x)= -2x^2 +48x, where x is the width of the area in meters and A(x)is the area in square meters. What is the maximum area that can be enclosed?
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A(x)= -2x^2 +48x
complete the square:
A(x)= -2(x^2 -24x+144)+288
A(x)=-2(x-12)^2+288
This is an equation of a parabola that opens down with vertex at (12,288)
maximum area that can be enclosed? =288 sq m