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A farmer is planning to put in a garden next to his barn. He is planning to fence in the garden on three sides (the barn will make the fourth side). He has 100 feet of fencing to use. Find an equation for the area of the garden. Explain/Show work.

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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L + 2W = 100   
 
L = 100 -2W
 
 
W ( 100 - 2W )     / area covered by fence
 
 
- 2w^2 + 100 W    / maximum area covered
 
 
 
f( w ) = - 2 w ^2 + 100w
 
 f( w ) is maximum value at  w = -100/ -4  = 25 ft
 
   Maximum area of garden is a square with side = 25 ft
 
 
 
 
 
 
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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A farmer is planning to put in a garden next to his barn. He is planning to fence in the garden on three sides (the barn will make the fourth side). He has 100 feet of fencing to use. FIND AN EQUATION FOR THE AREA OF THE GARDEN. Explain/Show work.

x = width of garden

y = length of garden

Don’t have to use fence along barn, so:

2x + y = 100 ==> y = 100 - 2x

Area, A = xy = x(100 - 2x)

A(x) = 2x(50 - x) graphs as a parabola opening down.

So an answer to the question would be:
A(x) = 2x(50 - x), or
A(x) = 100x - 2x^2.

But let’s find the dimensions of garden with largest area.

Zeros of A(x) are x = 0 and x = 50.

Axis of Symmetry is half way between at x = 25.

Vertex is maximum point of A(x) and lies on AoS; so

Vertex = (25,A(25)).

y = 100 - 2(25) = 50

So largest garden is A RECTANGLE 25 ft x 50 ft,

and has area A(25) = 2(25)(50-25) = 1250 ft^2