Quadratic Functions Word Problem

AI Thread Summary
The problem involves maximizing the area of a rectangular enclosure with a budget of $2400 for fencing, where the high fence costs $8 per foot and the low fence costs $4 per foot. The cost equation is established as 16x + 20y = 2400, where x is the length and y is the height of the enclosure. To find the maximum area, the area function A = xy is derived, and calculus techniques, such as finding the vertex of the quadratic function, are suggested for optimization. The final dimensions that maximize the area are determined to be 75 feet by 60 feet. This approach effectively combines geometry and algebra to solve the problem.
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Homework Statement



Your budget for constructing a rectangular enclosure which consists of a high surrounding fence and a lower inside fence that divides the enclosure in half is 2400 dollars. The high fence costs 8 dollars per foot and the low fence costs 4 dollars per foot. Find the dimensions and the maximum area of each half of the enclosure.

Homework Equations



-None-

The Attempt at a Solution



I know you probably have to figure out how much each type of fencing is going to cost then divide that for the particular fence giving you the dimensions. I don't know how to set up the proper formula for getting it though. I tried 16x+8y=2400 then solving for y and plugging it back in but I keep going in circles. I couldn't get much further than that because I am confused.

Thanks for any help!
 
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First, draw a picture of this problem. Remember, you are trying to maximize area enclosed while staying under budget with the purchase of the fence. You are going to need to write an equation for the area enclosed by the fence and another equation for the cost of the fence.
 
there is a picture in the book but I still don't understand it

Area= xy
Cost of fence, 8x+4y=2400 ?

?

Image:

8ZYmfeM.png
 
Last edited:
Let x be length, y be height:

Low Fence = y.
2*8*x = Cost of Horizontal Fence
2*8*y + 4*y = Cost of Vertical Fence + Cost of Low Fence

16x + 20y = 2400

Thus, y = (2400-16x)/20

You might need calculus from here onward. Did you learn calculus?
 
I am in precalculus
After you solve for y do you plug it back into the original equation?
 
No, you shouldn't. I'm not very sure how to do this in this case, because I only know the Calculus method, sorry
 
The enclosed area A=xy must be maximum. Substitute for y: You get a quadratic function A(x). If you plot it, at what x is the peak of the parabola? (Use completing the square)

ehild
 
Thanks for the help I just used x=-b/2a to find h and subbed it back in for k giving me (h, k) the vertex which are the lengths 75' x 60' of the sides
 

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