Applied Maximum and Minimum Problems help

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hank

Junior Member
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Sep 13, 2006
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Here's the problem:

A field has a boundary a right triangle with hypotenuse along a straight stream. A fence bounds the other two sides of the field. Find the dimensions of the field with maximum area that can be enclosed using 1000' of fence.

I need to find an applicable formula here, then apply the derivative and find the max.
So I'm thinking the formula here has to be A = 1/2bh, but what about they pythagorean theorem? Does that play in here as well?

What about the fence? does b +h = 1000 play in there?

I'm sooo confused..
 
A field has a boundary a right triangle with hypotenuse along a straight stream. A fence bounds the other two sides of the field. Find the dimensions of the field with maximum area that can be enclosed using 1000' of fence.

I need to find an applicable formula here, then apply the derivative and find the max.
So I'm thinking the formula here has to be A = 1/2bh, but what about they pythagorean theorem? Does that play in here as well?

What about the fence? does b +h = 1000 play in there?
Click to expand...

You've got all the pieces; let's put them together:

A = (1/2)bh = (1/2)(1000-h)h = (1/2)(1000h - h^2)

That's a quadratic and will graph as an inverted parabola. The max A value is at the vertex of the parabola (where the derivative equals zero).
 
Hello, Hank!

A field has a boundary a right triangle with hypotenuse along a straight stream.
A fence bounds the other two sides of the field. .Find the dimensions
of the field with maximum area that can be enclosed using 1000' of fence.

I need to find an applicable formula here, then apply the derivative and find the max . . . Right!
So I'm thinking the formula here has to be A = 1/2bh . . . Correct!
But what about the Pythagorean theorem?
.Does that play in here as well? . . . not needed

What about the fence? .Does \(\displaystyle b\,+\,h\:=\:1000\) play in there? . . . Yes!
Click to expand...

You already have the necessary relationships . . . just put them together!

You have: \(\displaystyle \:b\,+\,h \:=\:1000\;\;\Rightarrow\;\;h\:=\:1000\,-\,b\;\) [1]

And you have: \(\displaystyle \,A \:=\:\frac{1}{2}bh\;\) [2]

Substitute [1] into [2]: \(\displaystyle \:A \:=\:\frac{1}{2}b(1000\,-\,b)\;=\;500x\,-\,\frac{1}{2}x^2\)

Go for it!

 
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