SOLUTION: Studying quadratic equations. Can you help with an equation for this word problem? Laura wants to plant a rectangular garden next to her house. she needs to enclose the garden

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Question 817049: Studying quadratic equations. Can you help with an equation for this word problem?
Laura wants to plant a rectangular garden next to her house. she needs to enclose the garden with a fence, and will use the house as one side of the fence. If laura has 63 feet of fencing to use, what is the maximum area that she can enclose? What are the dimensions of the garden?

Found 2 solutions by TimothyLamb, josgarithmetic:
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
63 = 2w + L
L = 63 - 2w
a = wL
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a = wL
a = w(63 - 2w)
a = 63w - 2w^2
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a(w) = -2w^2 + 63w
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the above quadratic equation is in standard form, with a=-2, b=63, and c=0
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to solve the quadratic equation, plug this:
-2 63 0
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a maximum at ( w= 15.75, a(w)= 496.125 )
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Answer:
the maximum area is 496.125 sq.ft ( a(w) from the vertex )
w = 15.75 ft ( w from the vertex )
L = 31.50 ft ( calculated from above equation for L using w )
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Solve quadratic equations, quadratic formula:
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Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = a dimension and y = the other dimension.
A is for area.
A=xy and 2x+y=63; this second equation is for the fencing length.
You want the maximum A and this you might guess if you look at the two equations, would be a quadratic function. You can see I am using y as the length of the enclosure which is against the house.

y=-2x%2B63 is from the fencing length equation;
A=x%2A%28-2x%2B63%29
A=-2x%5E2%2B63x
!.................That function is a parabola, opening downward, so it has a maximum point. You can find the roots (the values of x for which A is zero) and the x value in the exact middle of those roots will occur at the maximum point of A.

Here we go: A=0=-2x%5E2%2B63x
0=x%28-2x%2B63%29
One possibility for x is highlight_green%28x=0%29.
Other possibility for x is -2x%2B63=0
2x=63
x=63%2F2
highlight_green%28x=31%261%2F2=63%2F2%29
'
The x value in the middle: %2863%2F2%29%2F2=highlight%2863%2F4=15%263%2F4%29

Going back to the formula for y derived from the fence length above, you should find that y and x are the same value; so the maximum area is a square.