SOLUTION: Please help me- I cannot get this I have tried to get you help a few times on this. I am getting nowhere. please please help. Thank you Suppose you wanted to constr

Question 90316: Please help me- I cannot get this I have tried to get you help a few times on this. I am getting nowhere. please please help. Thank you

Suppose you wanted to construct a fence around a garden plot in the form of a rectangle. On the neighbor�s side it�s going to need heavy-duty fencing that costs $2.00 per foot. The other three sides can be made of standard fencing material that costs $1.20 foot. You have $200 to spend.
1. How many different rectangles would it be possible to enclose for $200?
2. Write an equation using two variables for the total cost of the fence. Use x and y to represent the width and length of the rectangle. Solve the equation for one of the variables (either x or y).

3. Given that Area = (x)(y), now write the area function for this problem using ONLY ONE VARIABLE (by making a substitution using the equation from #2).

4. What dimensions will give the maximum area inside the fence?
5. What will be the maximum area?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose you wanted to construct a fence around a garden plot in the form of a rectangle. On the neighbor�s side it�s going to need heavy-duty fencing that costs $2.00 per foot. The other three sides can be made of standard fencing material that costs $1.20 foot. You have $200 to spend.
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Draw the rectangle; let the fence on the neighbors side have length x.
The side opposite it is also of length x.
Let the other two sides have length y.
1. How many different rectangles would it be possible to enclose for $200?
Plenty
-----------------
2. Write an equation using two variables for the total cost of the fence. Use x and y to represent the width and length of the rectangle. Solve the equation for one of the variables (either x or y).
Cost Equation: 2x + 1.2x + 1.2y + 1.2y = 200
3.2x + 2.4y = 200
y = (-3.2/2.4)x + (200/2.4)
y = (-4/3)x + (250/3)
----------------
3. Given that Area = (x)(y), now write the area function for this problem using ONLY ONE VARIABLE (by making a substitution using the equation from #2).
Area = x[(-4/3)x + (250/3)] = -(4/3)x^2+(250/3)x
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4. What dimensions will give the maximum area inside the fence?
Maximum area occurs at x=-b/2a = -(250/3)/[2(-4/3)] = 31.25 ft (width)
Then y = (-4/3)*(125/4)+(250/3) = -125/3 + 250/3 = 125/3 = 41.66 ft (length)
--------------------
5. What will be the maximum area?
Area = x*y = 31.25*41.66 = 1301.85 sq ft.
=====================
Cheers,
Stan H.

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