HW11-solutions

# HW11-solutions - white(taw933 HW11 ben-zvi(55600 This...

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white (taw933) – HW11 – ben-zvi – (55600) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Arectangulardogpoundwiththreekennels as shown in the fgure consists oF a rectangular Fenced area divided by two partitions. Determine the maximum possible area oF this pound iF 72 yards oF chain link Fencing is available For its construction. 1. max area = 164 sq.yards 2. max area = 165 sq.yards 3. max area = 163 sq.yards 4. max area = 162 sq.yards correct 5. max area = 161 sq.yards Explanation: Let the dimensions oF the ±oor oF the dog pound be as shown in the fgure below x y Then the area oF the pound is given by A = xy , while the total Fencing needed is the sum oF the perimeter 2 x +2 y and the inner partitions 2 y .S ince72ya rd so FFenc ingava i lab lew ege t arelation 72 = 2 x +4 y i.e. ,36 = x +2 y .E l im i n a t i n g y From these two equations gives an expression A = 1 2 x (36 - x )=1 8 x - 1 2 x 2 For the area solely as a Function oF x .A sth e maximum value oF x is x =36 ,themax imum area will thus be the absolute maximum value oF A on the interval [0 , 36]. This maximum will occur at a critical point x 0 oF A in (0 , 36) or at an endpoint. Now A ± ( x )=1 8 - x =0 when x 0 =18.But A (0) = 0 ,A ( x 0 )=1 6 2 ,A (36) = 0 . Thus the maximum area oF the dog pound is max area = 162 sq.yards and this occurs when the pound is 18 yards wide. 002 10.0 points A6 ± × 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,

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white (taw933) – HW11 – ben-zvi – (55600) 2 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 . DiFerentiating 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 =16cu .ins . . 003 10.0 points Ah om e own e
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HW11-solutions - white(taw933 HW11 ben-zvi(55600 This...

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