Question 1114192: 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
100 ft of fence? What should the dimensions of the garden be to give this area?
The maximum area that the farmer can enclose with 100 ft of fence is______. There might be more to this problem, but right now, this is all I have. Thank you.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this is what i get.
let the length of the side of the barn be L, since no length was given.
you have 100 feet of fence.
that would form the perimeter of the garden, with the side of the barn being one side of the garden.
the perimeter of a rectangle is 2L +2Y, where L is the length and W is the width.
since the barn is already there, then you only need one L and two 2W to form the perimeter of the garden that is composed of the fence.
you therefore have the perimeter being equal to L + 2W.
the formula for the perimeter is therefore 100 = L + 2W.
the formula for the area of the garden is L * W, where L is the length and W is the width.
you have 2 equations that need to be solved simultaneously.
they are:
L + 2W = 100
L * W = area
solve for W in the first equation to get W = (100 - L)/2.
replace W in the second equation to get L * (100 - L)/2 = area.
simplify this equation to get (100L - L^2) / 2 = area.
simplify this equation further and reorder the terms in descending order of degree to get:
area = -.5*L^2 + 50*L
this is a quadratic equation that can graphed.
if you let y = area and x = L, then the equation becomes:
y = -.5x^2 + 50x
if you set y = 0, this equation is in standard form of -.5x^2 + 50x = 0.
in this form:
a = coefficient of x^2 term = -.5
b = coefficient of x term = 50
c = constant term = 0.
the maximum value of this equation is when x = -b/2a.
this makes x = 50.
the value of y is equal to -.5 * 50^2 + 50 * 50 = 1250 when x = 50
since x represents L and y represents the area, you get L * W = area becomes 50 * W = 1250.
solve for W to get W = 25.
the area of the garden is maximized when L = 50 and W = 25.
the perimeter of the garden that is composed of fence = L + 2W = 50 + 2 * 25 = 100
the area is 50 * 25 = 1250.
the graph of the equation of y = -.5x^2 + 50x shows this to be true.
remember that x represents L, which is the length of the garden where one side of the length is the barn and the other side of the length is part of the fence.
here's the graph.
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