SOLUTION: Jeremiah is going to enclose a portion of his backyard with fence for his dogs. He has 48 metres of fence. The area that can be enclosed by the fence is modelled by the function A(

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Question 1038661: Jeremiah is going to enclose a portion of his backyard with fence for his dogs. He has 48 metres of fence. The area that can be enclosed by the fence is modelled by the function A(x)=-2x^2+48x where x is the width of area in metres and A(x) is the area in square metres. What is the maximum area that can be enclosed?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A%28x%29=-2x%5E2%2B48x=%28-2x%2B48%29x

The width is x and the length is -2x+48.

The description and question are not really completely clear. The 48 meters of fence could be for four sides or could be for three sides using the boundary of the backyard as one of the sides. Also, you want to maximize A using the variable, x.

dA%2Fdx=-4x%2B48
Maximized for -4x%2B48=0
4x=48
x=12---------the width.

Now use the width to find length.
-2x%2B48
-2%2812%29%2B48
48-24
24-----------Length. ... Must be wrong.

Try differently.
Forty Eight meters of fence material, assuming to be used for all four sides.
x and y dimensions.
2x%2B2y=48
x%2By=24
x=24-y
or
y=24-x
highlight_green%28A=xy%29, A for area.
A=x%2824-x%29
But this pathway to solve, is not really according to the SPECIFIED MODEL TO USE.

Going back to the specified model then means the expected answer, again assuming all four sides are fenced using the given length of material,
x=12
and the area will need to be
-2%2A%2812%29%5E2%2B48%2A12
-288%2B48%2A12
highlight%28576%29 square meters.