SOLUTION: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the       Log On


   

Question 622246: Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the side of the warehouse as one side of the rectangular fenced area. What are the dimensions of the maximum area Randy can enclose? Use an algebraic method to solve.
Found 2 solutions by math-vortex, jsmallt9:
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--
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Here is one way to solve this problem.
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Let L be the length of the fence enclosure.
Let W be the width of the fence enclosure.
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Let's have the side of the warehouse be one of the lengths. (This is arbitrary. We could choose to have it be one of the widths and the still work the problem.)
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We have 120m of fencing material. From this 120m we have two widths and one length. Just to be clear, Randy builds a rectangular fence enclosure. One length of the rectangle is the side of the warehouse, L meters. The other length of the rectangle is made of fencing material, L meters. The two widths of the rectangle are made of fencing material, W meters each.
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We can write an equation for the fencing material:
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[one length] + [two widths] = 120
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L%2B2W=120
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We can also write an equation for the area of the fenced enclosure. The formula for the area A of a rectangle is
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A=L%2AW
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We have have two equations with the variables L and W. We will use the substitution method to begin solving this system.
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Rewrite the first equation to give L in terms of W.
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L%2B2W=120
L=120-2W
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Now substitute 120-2W for L in the second equation.
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A=L%2AW
A=%28120-2W%29%2AW
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Simplify.
A=120W-2W%5E2
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Rewrite the equation to show the terms in descending order.
A=-2W%5E2%2B120W
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Notice that we have a quadratic equation. The graph of a quadratic is a parabola.
When the leading coefficient of the quadratic--the coefficient of the W-squared term--is negative, the parabola opens downward and the the vertex is the maximum point.
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The leading coefficient is negative in this problem; the coefficient of the the W-squared term is -2.
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This means that the area is maximized at the vertex. When we find the point (W,A) at the vertex, we will know the area A of the fenced enclosure and the width W of that fence.
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To find the vertex, we need to have our quadratic equation in general form,
y=ax%5E2%2Bbx%2Bc
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General form of our equation is A=-2W%5E2%2B120W%2B0, so a=-2, b=120, and c=0.
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The x-coordinate of the vertex is the value -b/2a. So, at the vertex,
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W=-%28b%29%2F2a
W=-%28120%29%2F%28%282%29%28-2%29%29
W=-%28120%29%2F%28-4%29
W=30
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So the width is 30m. Now we substitute 30 for W in the Area equation.
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A=-2W%5E2%2B120W
A=-2%2830%29%5E2%2B120%2830%29
A=-1800%2B3600
A=1800
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The maximum area we can enclose is 1800m. Since the area is 1800m and the width is 30m, the length of the enclosure must be 60m since 30*60=1800.
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I hope this helps. PLease email me if I need to explain any part better.
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Ms.Figgy
math.in.the.vortex@gmail.com

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
It will probably help to have a diagram to look at:
  1. Since we're told that the fenced area is to be a rectangle, draw a rectangle.
  2. Since one side is to be the barn, choose a side of the rectangle to be a the barn. (It doesn't matter which side you choose.)
  3. Label the two sides of the rectangle that are connected to the barn as "x". (Since this is a rectangle these opposite sides should be the same.)
  4. The hardest part is the side opposite the barn. We know that we have 120 meters of fence to use. It hope it makes sense that to get the maximum area we should use all the fencing. So the lengths of the three "non-barn" sides will be 120. So how do we express the length of the side opposite the barn? Well if we know what "x" was it would be easy. For example if x = 35 then the side opposite the barn would be 120 - 35 - 35 = 50. We now use this same logic even though we don't know what "x" is yet. The side opposite the barn will be
    120 - x - x
    which simplifies to:
    120 - 2x
    Label the side opposite the barn as 120-2x.
With this diagram we can now find a solution. The area of any rectangle is length times width (or base times height). So for our rectangle the area will be:
A+=+x%2A%28120-2x%29
which simplifies to:
A+=+120x-2x%5E2
This is a quadratic equation so I am going to rewrite it in ax%5E2+%2B+bx+%2B+c form:
A+=+-2x%5E2+%2B+120x

We should recognize that the graph of a quadratic equation will be a parabola. And since the squared term is x%5E2 this parabola will open upward or downward. And since the "a" is negative (-2 to be specific), this parabola will open downward. If we picture a downward-opening parabola in our minds we should be able to understand that the vertex of the parabola will be the highest point of the graph. So the x-coordinate of the vertex will give the the maximum value for A, the area.

So we just need to figure out the x-coordinate of the vertex. This can be done by completing the square or by just knowing that the x-coordinate of the vertex would be -b%2F2a. Our "b" is 120 and our "a" is -2 so the x-coordinate of the will be:
-%28120%2F%282%2A%28-2%29%29%29+=+30

This is the x that gives us the maximum area. (If you're curious as to exactly what number this maximum area is, just substitute 30 for x into A+=+-2x%5E2+%2B+120x)

So we can label the two sides of "x" as 30's. And since the side opposite the barn is 120 - 2x, it will be 120 - 2(30) = 120 - 60 = 60 meters.