A farmer wants to construct a rectangular pen for his pigs using the side of his barn as one side of the pen and 120 feet of fencing for the other 3 sides. In order to make the area inside the pen as large as possible, what should be the dimensions of the pen?
In your subdivision you have an area planted with wildflowers and trails to walk. To keep the area natural, you want to enclose the largest rectangular area possible with 2000 feet of fencing. Write an equation for the area A.
200 yards of fencing is to be used to build a rectangular enclosure with 3 equal parts. What is the maximum area that be enclosed?
A go-cart track has about 380 racers per week and charges $35 to race. The owner estimates that there will be 20 more racers per week for evey $1.00 reduction in the price per ticket. How can the go-cart owner maximize weekly revenue?
can u please tell me what the equations are?
Let x = width of pigpen
y = length of pigpen
Perimeter = 2x + 2y
Area = xy
The length of the 4th side is unknown, but the 3 sides to be built sum to 120:
120 = 2x + y (or 120 = x + 2y)
so
120 - 2x = y
and
A(x) = x(120 - 2x)
A(x) = 120x - 2x²
or in vertex form for parabola:
A(x) = -2(x² - 60x)
A(x) = -2(x² - 60x + 900) + 1800
A(x) = -2(x - 30)² + 1800
Which is a parabola with vertex maximum at (30, 1800)
so pigpen is 30 x 60, area = 1800
Or solve for 1st derivative:
A'(x) = 120 - 4x
So when m = 0:
0 = 120 - 4x
x = 30
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Likewise
Perimeter of garden = 2x + 2y
Area = xy
So
2000 = 2x + 2y
2000 - 2x = 2y
1000 - x = y
so
A(x) = x(1000 - x) = 1000x - x²
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I don't understand what "rectangle with 3 equal parts" is supposed to mean. Three squares, three rectangles or ?
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Let x = charge to race
y = # of racers
Given:
(35, 380)
(34, 400)
400 - 380 = m(34 - 35)
20 = -m
-20 = m
y - 400 = -20(x - 34)
y = -20x + 1080
Revenue = xy
R(x) = x(1080 -20x)
Taylor Dunn elctric golf cart truck with enclosure
