Solutions HW 11

# Solutions HW 11 - mao(tm23477 – Hw 11 – gualdani...

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Unformatted text preview: mao (tm23477) – Hw 11 – gualdani – (57180) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the positive number such that the sum of 3 times this number and 8 times its reciprocal is as small as possible. Correct answer: 1 . 63299. Explanation: Let x be a positive number. Then we have to find which x in (0 , ∞ ) minimizes the func- tion f ( x ) = 3 x + 8 x . Now f ′ ( x ) = 3- 8 x 2 , f ′′ ( x ) = 16 x 3 . Thus f is concave up everywhere on (0 , ∞ ) and x = radicalbigg 8 3 is the critical point of f for which f ( x ) is the absolute minimum value of f on (0 , ∞ ) . Consequently, x = radicalbigg 8 3 = 1 . 63299 ≈ 1 . 63299 . 002 10.0 points Find the height of the largest rectangle that can be inscribed in the region bounded by the x-axis and the graph of y = radicalbig 49- x 2 . Correct answer: 4 . 94975 units. Explanation: Since the graph is upper half of the circle x 2 + y 2 = 49 having radius 7 and center at the origin, the rectangle is the one shown in P ( x,y ) 7- 7 7 Thus the area of the rectangle is A = 2 xy = 2 y radicalbig 49- y 2 , so we have to maximize A ( y ) on the interval [0 , 7]. Now A ′ ( y ) = 2 radicalbig 49- y 2- 2 y 2 radicalbig 49- y 2 = 2(49- 2 y 2 ) radicalbig 49- y 2 . Thus the critical points of A ( y ) occur at y =- 7 √ 2 , 7 √ 2 , only one of which lies in [0 , 7]. But A (0) = 0 , A parenleftBig 7 √ 2 parenrightBig = 49 , A (7) = 0 . Consequently, the maximum area = 49 sq. units , and this occurs when the rectangle has height = 7 √ 2 = 4 . 94975 . 003 10.0 points A farmer wishes to erect a fence enclosing a rectangular area adjacent to a barn which is 16 feet long. His plan mao (tm23477) – Hw 11 – gualdani – (57180) 2 Fenced area 16 ft Barn for the fenced area indicates the fencing in bold black lines; it shows also that the barn will be used as part of the fencing on one side. Find the largest area, A max , that can be enclosed if 96 feet of fencing material is available. 1. A max = 780 sq. ft. 2. A max = 776 sq. ft. 3. A max = 782 sq. ft. 4. A max = 778 sq. ft. 5. A max = 784 sq. ft. correct Explanation: Let x, y be the respective parts of the fence shown in the figure 16 ft x y Barn Then 2 x + 16 + 2 y = 96 , so adjacent sides of the fenced area will have length x + 16 , 40- x. Hence, as a function of x , the area enclosed by the fence will be given by A ( x ) = (40- x )( x + 16) . Differentiating A ( x ) with respect to x we see that A ′ ( x ) =- 2 x + 24 . The critical points of A are thus the solutions of A ′ ( x ) = 24- 2 x = 0 , i.e. x = 12. On practical grounds (or use the fact that A ′′ =- 2), this will yield a maximum value of A ( x ). At x = 12, therefore, the enclosed area will have a maximum value A max = 784 sq. ft....
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## This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.

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Solutions HW 11 - mao(tm23477 – Hw 11 – gualdani...

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