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18_Cal_Solution of Calculus_6e

# 18_Cal_Solution of Calculus_6e - C01S02.066 Recall that the...

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C01S02.066: Recall that the area of the rectangle is given by y = A ( x ) = x (50 x ). To maximize A ( x ) we find the vertex of the parabola: y = 50 x x 2 = ( x 2 50 x ) = ( x 2 50 x +625 625) = ( x 25) 2 +625. Because the vertex of the parabola is at (25 , 625) and x = 25 is in the domain of the function A , the maximum value of A ( x ) occurs at x = 25 and is A (25) = 625 (ft 2 ). C01S02.067: If two positive numbers x and y have sum 50, then y = 50 x and x < 50 (because y > 0). To maximize their product p ( x ) we find the vertex of the parabola y = p ( x ) = x (50 x ) = ( x 2 50 x ) = ( x 2 50 x + 625 625) = ( x 25) 2 + 625 , which is at (25 , 625). Because 0 < 25 < 50, x = 25 is in the domain of the product function p ( x ) = x (50 x ), and hence the maximum value of the product of x and y is p (25) = 625. C01S02.068: Recall that if x new wells are drilled, then the resulting total production p is given by p ( x ) = 4000 + 100 x 5 x 2 . To maximize p ( x ) we find the vertex of the parabola y = p ( x ) = 5 x 2 + 100 x + 4000 = 5( x 2 20 x 800) = 5( x 2 20 x + 100 900) = 5( x 10) 2 + 4500 .
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