SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 236 ft. What dimensions would yield? The length that would yield the maximum area is _

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Question 408213: A carpenter is building a rectangular room with a fixed perimeter of 236 ft. What dimensions would yield?
The length that would yield the maximum area is __ft?
Answer by katealdridge(100) (Show Source):
You can put this solution on YOUR website!
Area is the function you want to maximize.
A=L*W
Perimeter is the function that limits your area and is also a tool to eliminate one of the variables.
P=2L+2W=236 Solve this equation for either L or W. We'll solve for W.
2W=236-2L subtracting 2L from both sides
W=118-L dividing both sides by 2
Now substitute this into the area equation:

distributing the L
This is a quadratic function which graphs as a parabola. You need to find the vertex of the parabola. Depending on how your teacher wants you to do that, is what you do next.
You could graph it on a graphing calculator and find the vertex that way, or you could do it by hand. It also kind of depends on what class you're in. You could do it different ways in algebra vs calculus. This is how you would do it algebra:
The x coordinate (in this case we're using L) of the vertex is found by the formula:
when the quadratic is in Standard Form
this would be in standard form, with a = -1, b = 118
This is the length at which the area is maximized.
You could then substitute the 59 in for L in the above equations to find the width or the area.

SOLUTION: A carpenter is building a rectangular room with a fixed perimeter of 236 ft. What dimensions would yield? The length that would yield the maximum area is __ft?