Quest HW 11-solutions - syed(sms3768 – Quest HW 11 –...

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Unformatted text preview: syed (sms3768) – Quest HW 11 – seckin – (56425) 1 This print-out should have 16 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 4 times its reciprocal is as small as possible. Correct answer: 1 . 1547. Explanation: Let x be a positive number. Then we have to find which x in (0 , ∞ ) minimizes the func- tion f ( x ) = 3 x + 4 x . Now f ′ ( x ) = 3- 4 x 2 , f ′′ ( x ) = 8 x 3 . Thus f is concave up everywhere on (0 , ∞ ) and x = radicalbigg 4 3 is the critical point of f for which f ( x ) is the absolute minimum value of f on (0 , ∞ ) . Consequently, x = radicalbigg 4 3 = 1 . 1547 ≈ 1 . 1547 . 002 10.0 points A 6 ′′ × 6 ′′ square sheet of metal is made into an open box by cutting out a square at each corner and then folding up the four sides. Determine the maximum volume, V max , of the box. 1. V max = 16 cu. ins. correct 2. V max = 26 cu. ins. 3. V max = 31 cu. ins. 4. V max = 36 cu. ins. 5. V max = 21 cu. ins. Explanation: Let x to be the length of the side of the squares cut from each edge. Then the volume of the resulting box is given by V ( x ) = x (6- 2 x ) 2 . Differentiating V with respect to x we see that dV dx = (6- 2 x ) 2- 4 x (6- 2 x ) . The critical points of V are thus the solutions of 3 x 2- 12 x + 9 = 0 , i.e. , x 1 = 1 , x 2 = 3 , where the second one can be disregarded for practical reasons. At x = x 1 , therefore, V ( x ) becomes V max = 16 cu. ins. . 003 10.0 points The rectangle in the figure b ( x, y ) is formed with adjacent sides on the coordi- nate axes and one corner on the graph of y = 10 x + 24 x 2 + 1 Find the maximum possible area of this rect- angle. syed (sms3768) – Quest HW 11 – seckin – (56425) 2 1. max area = 37 2 sq. units. 2. max area = 39 2 sq. units. 3. max area = 19 sq. units. 4. max area = 18 sq. units. correct 5. max area = 35 2 sq. units. Explanation: The rectangle has side lengths x and y , so its area is given by A = xy . On the other hand, one corner lies on the graph of y = 10 x + 24 x 2 + 1 . Thus the area of the rectangle can expressed solely as a function of x by A ( x ) = x parenleftBig 10 x + 24 x 2 + 1 parenrightBig = 10 x 2 + 24 x x 2 + 1 . Differentiating with respect to x we see that A ′ ( x ) = (20 x + 24)( x 2 + 1)- 2 x (10 x 2 + 24 x ) ( x 2 + 1) 2 = 24 + 20 x- 24 x 2 ( x 2 + 1) 2 . At a maximum A ′ ( x ) = 0, i.e. , 24 + 20 x- 24 x 2 ( x 2 + 1) 2 = 0 , so the points x 1 , x 2 at which A ′ ( x ) = 0 are the solutions of 24 x 2- 20 x- 24 = (4 x- 6)(6 x + 4) = 0; in other words, x 1 = 3 2 , x 2 =- 2 3 . On practical grounds the positive solution corresponds to maximum area, while the neg- ative solution has no meaning for this prob- lem. Consequently, the maximum area = 18 sq. units. , which is achieved at x = 3 2 ....
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Quest HW 11-solutions - syed(sms3768 – Quest HW 11 –...

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