1. Maximized

Hello! I am working on a math problem and I can't seem to do it at all. Any help would be greatly apprieciated!! Sorry if the picture is big, but I wanted to draw it to give what's on the paper.

A Farmer has 400 feet of fence and wants to construct a triangular pen using his barn as the boundary for one end of the pen (see figure below - note that no fencing is used along the side against the barn). Use the figure below as well as the information given above to anser the questiosn that follow.

a) Find the function hyp(B), the length of the pen's hypotenuse as a function of the length of its base.

b) Fidn the function A(B), the area of the pen as a function of the length of it's base. You do not need to simplify your answer.

c) Sketch the graph of the function y=A(B). Label any important points.

d) Use your answers from part a-c to determine the dimensions of the pen with maximum area. Round your answers to the nearest tenth of a foot.

base = _____________
height = _____________
hypotenuse = ___________

Again thanks for any help!!!

2. Originally Posted by antz215
Hello! I am working on a math problem and I can't seem to do it at all. Any help would be greatly apprieciated!! Sorry if the picture is big, but I wanted to draw it to give what's on the paper.

A Farmer has 400 feet of fence and wants to construct a triangular pen using his barn as the boundary for one end of the pen (see figure below - note that no fencing is used along the side against the barn). Use the figure below as well as the information given above to anser the questiosn that follow.

a) Find the function hyp(B), the length of the pen's hypotenuse as a function of the length of its base.

As there are 400 ft of fence we have:

Hyp + B =400

So:

Hyp = 400 - B.

b) Fidn the function A(B), the area of the pen as a function of the length of it's base. You do not need to simplify your answer.
Let the length of the barn side of the enclosure be x, then

A(B) = B*x/2,

and

Hyp^2 = x^2 + B^2

or substituting Hyp from part a):

(400-B)^2 = x^2 + B^2

so:

x^2 = (400-B)^2 - B^2 = 160000 -800 B

Plug into the area formula:

A(B) = (1/2) B sqrt(160000 - 800 B)

RonL

3. Thanks sooooo much!!!!

If anyone (or you incase you just didn't see it) has help on parts c & d it'd be greatly appriciated!!!

4. Originally Posted by antz215
Hello! I am working on a math problem and I can't seem to do it at all. Any help would be greatly apprieciated!! Sorry if the picture is big, but I wanted to draw it to give what's on the paper.

A Farmer has 400 feet of fence and wants to construct a triangular pen using his barn as the boundary for one end of the pen (see figure below - note that no fencing is used along the side against the barn). Use the figure below as well as the information given above to anser the questiosn that follow....

Hi,

to a) B and the hypotenuse have together a length of 400 '. So your function is:
hyp(B) = H = 400 – B

H must be longer than B. So the domain of the function is 0 < B < 200

to b) A right triangle is the half of a rectangle. One side of this rectangle is B. Use the Pythagorean formula to calculate the missing side L:

L² = H² – B²
L² = (400 – B)² – B² = 160000 – 800B+ B² – B² = 160000 – 800B. Thus

L = sqrt( 160000 – 800B )

Therefore the area is: A(B) = ½ * B * L

A(B) = ½ * B * sqrt(160000 – 800B) and 0 < B < 200

I've attached a diagram.

(Hello, CaptBlack, you are much too fast for me!)

5. Originally Posted by earboth
(Hello, CaptBlack, you are much too fast for me!)
I posted ~7.5hrs before you

I expect you didn't notce my reply until after you posted. I do
that all the time

RonL