Respuesta :
L + 2w = 120
L = 120 - 2w
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A = Lw
A = (120 - 2w)w
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Area is now a quadratic function of width (w):
A(w) = -2ww + 120w + 0
A(w) = (-2)w^2 + 120w + 0
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the above quadratic equation is in standard form, with a=-2, b=120, and c=0
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to find the maximum area A(w), plug this:
-2 120 0
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Answer 1:
the maximum point of the above quadratic equation is: ( 30, 1800 )
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so the maximum area that can be enclosed is: 1800 sq.m (the y-coordinate of the maximum point)
L = 120 - 2w
---
A = Lw
A = (120 - 2w)w
---
Area is now a quadratic function of width (w):
A(w) = -2ww + 120w + 0
A(w) = (-2)w^2 + 120w + 0
---
the above quadratic equation is in standard form, with a=-2, b=120, and c=0
---
to find the maximum area A(w), plug this:
-2 120 0
---
Answer 1:
the maximum point of the above quadratic equation is: ( 30, 1800 )
---
so the maximum area that can be enclosed is: 1800 sq.m (the y-coordinate of the maximum point)
Answer:
a(x) = x*y = x*(280 - 2x) square meters
Step-by-step explanation:
In the figure attached, a plot of the problem is shown. Let's call y the length of the side that parallels the river
The perimeter that the fence cover is:
2x + y = 280 meters
Isolating y:
y = 280 meters - 2x
The area of the field is:
a(x) = x*y = x*(280 - 2x) square meters
