Quest HW 11-solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online