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Differentiation question about 3-sides / Max Area watch

1. I was reading this question which states:

A farmer wants to make a rectangular corral along the side of a large barn and has enough materials for 60 meters of fencing. Only three sides must be fenced, since the barn wall will form the fourth side. What width of rectangle should the farmer use so that the maximum area is enclosed?

Two sides of the fence will be of length x; the third will be 60-2x. So the area will be x(60-2x) = 60x - 2x^2. To find the maximum area, find the first derivative and solve for zero.

y = 60x - 2x^2
y' = 60 - 4x
0 = 60 - 4x
x = 15

Technically, the roots of the first derivative can be either a maximum or a minimum. So let's check to be sure it's a maximum by evaluating y on either side of x = 15

x=14: y = 60(14) - 2(14)^2 = 448
x=15: y = 60(15) - 2(15)^2 = 450
x=16: y = 60(16) - 2(16)^2 = 448

So the area is maximized if x = 15, which means the two equal sides of the rectangle are 15 and the third side is 30.
I can understand the process given in the answer, though I'm wondering why the process of differentiation and solving for zero provides the required answer?

Thanks
2. Your quadratic is one for area, when you differentiate what you are finding is the rate of change of area with respect to x. If you find when that rate of change is 0, then your area is at a maximum or minium. In this case the maximum. So that is why solving for 0 finds the required answer.
3. (Original post by little pixie)
I was reading this question which states:

I can understand the process given in the answer, though I'm wondering why the process of differentiation and solving for zero provides the required answer?

Thanks
you have to imagine that the area is a curve. To find the maximum, you find where the gradient = 0 or y' = 0. this is a turning point where all other points on the curve are less than it.
this will give you either a minima or a maxima: a value of x that provides the biggest or smallest value of y.
4. Thanks very much for that. I think I understand it now. Thanks

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