Question 178406: A farmer decides to enclose a rectangular garden using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 100ft of fence? What should the dimensions of the garden be to give this area?
Answer by EMStelley(208)   (Show Source):
You can put this solution on YOUR website! Let's pretend that the barn side is one of the "lengths". So the perimeter of fencing would be 2 times the width plus one length. Let's call the width w and the length l. Then the amount of fencing needed is
We were told in the problem that there is 100 ft of fencing available, so we have
The question is to maximize area, which is A=lw. Now, currently we have too many variables to be able to solve anything. Can we fix this? Yes, using our perimeter equation, we can solve for one of the variables in terms of the other. Let's try solving for l (length). Then
Now we can replace l with this new expression in the area equation,
Ok, now we're down to only one variable, which is good. Let's distribute through on the right hand side so we can get a better idea of what we are looking at:
So, we have a quadratic equation. Notice, that since the coefficient of the w^2 term is negative, it's upside down. Just to get a visual, I will graph it for you.
So where is the maximum on this graph? It's the vertex. Now, to find the x coordinate of a quadratic equation of the form ax^2+bx+c, you have to calculate -b/2a. In this problem, a=-2 and b=100. So the x-coordinate (w) of the vertex is -100/(2(-2))=25. To find the y-coordinate (A), you plug in the x-coordinate back into the equation:
So the vertex is (25,1250).
This means that the maximum area is 1250 ft^2 and the dimensions giving this area are a width of 25 ft and a length of 50.
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