SOLUTION: I have a couple of homework problems that I am having trouble with. Any help you can give me to understand how to work these problems would be appreciated. The Cotes have 30

Question 168547: I have a couple of homework problems that I am having trouble with. Any help you can give me to understand how to work these problems would be appreciated.
The Cotes have 30 feet of picket fence with which to enclose a flower garden. What dimensions should the garden have in order to maximaze area? I believe the equation I would use is A(w) = (30-2)w.
Find the value of "a" such that f(x)=ax2+16x+38 has a minimum value of 6. Would this answer be 96/19?
Thank you.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I have a couple of homework problems that I am having trouble with. Any help you can give me to understand how to work these problems would be appreciated.
The Cotes have 30 feet of picket fence with which to enclose a flower garden. What dimensions should the garden have in order to maximaze area? I believe the equation I would use is A(w) = (30-2)w.
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30 feet is the perimeter of the enclosed area. The max area is a circle for a given perimeter. You didn't specify the shape. If it's to be a rectangle, the max area is a square.
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Find the value of "a" such that f(x)=ax2+16x+38 has a minimum value of 6. Would this answer be 96/19?
------------------
The minimum is the point where the 1st derivative is 0, so
2ax + 16 = 0
x = -8/a at the minimum.
Subbing -8/a for x and setting = 6:
a*(64/a^2) - 128/a + 38 = 6
64 - 128 + 38a = 6a
a = 2
Email me if you have questions, this is a complex problem. gsihoutx@aol.com

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