If 5300 feet of fencing are used to enclose a rectangular field, the resulting area of the field is A=x(2650-x) where x is the width of the pen. What is the maximum possible area of the pen?
A) 70,225
B)1,755,625
C)1325
D)33,125
Math Help?
2015-07-23 7:56 pm
回答 (3)
2015-07-23 9:00 pm
If 5300 feet of fencing is used and x=/=y then
5300=2x+2y, y=(5300-2x)/2, y=2650-x
A=xy, using y from above
A=x(2650-x)
A=2650x-x^2
dA/dx=2650-2x, d2A/dx2=-2, so that means that when dA/dx=0 it is an absolute maximum for A(x)
dA/dx=0 only when 2650=2x, x=1325
A(1325)=2650x-x^2=1,755,625
5300=2x+2y, y=(5300-2x)/2, y=2650-x
A=xy, using y from above
A=x(2650-x)
A=2650x-x^2
dA/dx=2650-2x, d2A/dx2=-2, so that means that when dA/dx=0 it is an absolute maximum for A(x)
dA/dx=0 only when 2650=2x, x=1325
A(1325)=2650x-x^2=1,755,625
2015-07-23 8:19 pm
A (x) = 2650 x - x²
A ` (x) = 2650 - 2x = 0 for max A
x = 1325 ft
A (1325) = 1325 [ 2650 - 1325 ] ft²
A (1325) = 1 755 625 ft²______Option B
A ` (x) = 2650 - 2x = 0 for max A
x = 1325 ft
A (1325) = 1325 [ 2650 - 1325 ] ft²
A (1325) = 1 755 625 ft²______Option B
2015-07-23 8:06 pm
The maximum area will be enclosed by a square pen. (5300 ft/4)^2 = 1,755,625 ft^2.
_____
The vertex of the expression for A is at x=2650/2.
Differentiating, you get dA/dx = 2650-2x. This is zero at x=2650/2.
_____
The vertex of the expression for A is at x=2650/2.
Differentiating, you get dA/dx = 2650-2x. This is zero at x=2650/2.
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