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padilas [110]
3 years ago
9

Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence

on that side. The other three sides will be enclosed with wire fencing. If Jacob has 400 feet of fencing, you can find the dimensions that maximize the area of the enclosure.
a) Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn). Write an function for the area A of the enclosure in terms of w . (HINT first write two equations with w and l and A . Solve for l in one equation and substitute for l in the other).

A(w) =

b) What width
w would maximize the area?
w = _____ ft
c) What is the maximum area?
A = _______square feet
Mathematics
1 answer:
Citrus2011 [14]3 years ago
6 0

Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn,so he needs no fence on that side.

Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).

one side of the length is not counted for perimeter because one side of length will be against the barn.

Perimeter = 400 ft

Perimeter of rectangle = L + W + W

400 = L + 2W

L = 400 - 2W

Area = L * W

Replace L by  400 - 2W

A(W) =  (400 - 2W) * W

A(W) = -2W^2 + 400W

Now we find out x coordinate of vertex to find the width that maximize the area

W = \frac{-b}{2a}

a= -2 and b = 400

W = \frac{-(400)}{2(-2)}=100

The width  w would maximize the area is w = 100ft

To find maximum area we plug in 100 for W in A(W)

A(W) = -2W^2 + 400W

A(W) = -2(100)^2 + 400(100)= 20000

the maximum area is 20,000 square feet

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The statement third "Betty wrote an indirect proof using contradiction' and statement fourth "Ken wrote direct proof using deductive evidence" are correct.

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3 0
1 year ago
Eli attended a concert. The decibel level of the music averaged 110 decibels but varied by 28 decibels from the average. Enter a
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Answer:

The decibel range is   82\leq d\leq 138

Step-by-step explanation:

Let

d -----> the decibel level of the music

we know that

The absolute value of the difference of the decibel level of the music and its average, must be less than or equal to 28 decibels

so

The absolute value that represent this problem is

\left|d-110\right|\le28

Solve the absolute value

<em>First solution (positive)</em>

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<em>Second solution (negative)</em>

-(d-110)\le28

Multiply by -1 both sides

(d-110)\ge-28

Adds 110 both sides

d\ge-28+110

d\ge82

therefore

The decibel range is

82\leq d\leq 138

8 0
3 years ago
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