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To solve this problem, we need an expression for the area in terms of one variable. Then we can use the first derivative to find the value of the variable which yields the maximum area.
First we must express the area in terms of the 2 unknowns:
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Now we must relate the dimensions to the fixed amount of fencing:
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We can combine these equations to express the area in terms of a single variable, x. Rewriting the fencing equation, we get
.
Now substitute in the equation for area and get

Now we find the derivative:
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If we set the derivative to 0, the condition at a maximum, we can solve for the x dimension, xMax:
,
or 
If we substitute this value in the above equation for y, we can obtain the corresponding yMax:
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We can find the maximum area by using these two dimensions in the original area equation:

We can check our result by trying values of x and y which depart from these slightly:
|
x |
90 |
110 |
|
y |
165 |
135 |
|
A |
44550 |
44550 |
The table reveals that xMax and yMax indeed result in a maximum area.