SOLUTION: Hi, "A gardener has 100m of fencing. He must use this to build the biggest fence possible (in square meters).There are 3 sides to the garden bed, attached to the wall which does n

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Question 811291: Hi,
"A gardener has 100m of fencing. He must use this to build the biggest fence possible (in square meters).There are 3 sides to the garden bed, attached to the wall which does not require fencing. What are the three side lengths you would use to get the biggest area of fencing?"
I know the answer is 50, 25, 25. But how do you develop an equation to solve this for full marks?
Thanks
Found 2 solutions by josgarithmetic, fcabanski:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Three sides are in lengths x, y, and y.
Two equations are possible.
Area, , and fence length, . The biggest area A is wanted.

The fence length equation is also , and we can solve for either variable and substitute into the A equation. Try ;
Then


That is a parabola with a maximum value.
Easiest to find the roots of the equation.


or , so right in between is . That is where A is maximum.
Answer by fcabanski(1391)   (Show Source): You can put this solution on YOUR website!
L + 2W = 100


A = LW


Find, from the first equation, one variable in terms of the other. L = 100-2W. Substitute that into the area equation.


A = (100-2W)W = 100W - 2W^2)


Use the first derivative to find the maximum possible area.


A' = 100 - 4W = 0


4W = 100 so W = 25


With these problems the value found when setting the derivative = to 0 us the max or min. But you can verify using the first derivative test.


0 <= W <= 50, so the ranges to check are 0 to 50 and 50 t0 100


W=1 is in the first range. Plug into the first derivative. 100- 4 = 96 which is positive.


w=50 is in the second range. 100-4(50) is negative.


The first derivative is positive before W=25 and negative after, so W=25 is a max.


L + 2(25) = 100 so L = 100-50 = 50.


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