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# Exam3 3 - 4 least cost = \$64 correct 5 least cost = \$60...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 3 and one corner at a point P ( x, y ) on the semi- circle as shown in P ( x,y ) 2 - 2 Then the area of the rectangle is given by A ( x ) = 2 xy = 2 x p 4 - x 2 . We have to maximize A ( x ) on the interval [0 , 2]. Now A 0 ( x ) = 2 p 4 - x 2 - 2 x 2 4 - x 2 = 2(4 - 2 x 2 ) 4 - x 2 . Thus the critical points of A ( x ) occur at x = - 2 2 , 2 2 , only one of which lies in [0 , 2]. But A (0) = 0 , A 2 2 · = 4 , A (2) = 0 . Consequently, max. area = 4 sq. units . keywords: optimization, semi-circle, critical point, maximum area, area rectangle 003 (part 1 of 1) 10 points A rancher wishes to build a fence to en- close a rectangular pen having area 32 square yards. Along one side the fence is to be made of heavy duty material costing \$6 per yard, while the remaining three sides are to be made of cheaper material costing \$2 per yard. Determine the least cost of fencing for the pen. 1. least cost = \$66 2. least cost = \$62 3. least cost = \$58
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Unformatted text preview: 4. least cost = \$64 correct 5. least cost = \$60 Explanation: Let x be the length of the side made of heavy duty material, and y the length of the adjacent sides. Then the cost of the fencing is given by C ( x,y ) = 6 x + 2( x + 2 y ) = 2(4 x + 2 y ) , while the area of the Feld is given by Area = xy = 32 . As a function solely of x , therefore, C ( x ) = 2 n 4 x + 64 x o . Di±erentiating C ( x ) with respect to x we see that C ( x ) = 2 n 4-64 x 2 o . The critical point of C is thus the solution of C ( x ) = 2 n 4-64 x 2 o = 0 , i.e. , x = ± 4. Since C 00 ( x ) = 128 x 3 , C ( x ) will be minimized when x = 4 (which is the only critical point to make practical sense anyway!). Consequently, the pen has least cost = \$64 , and this occurs when it is a 4 × 8 yard rectan-gle....
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