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Unformatted text preview: S olutions to M ath 118 F inal E xam G iven M ay 6, 2003 Problem 1. A manufacturer’s cost in dollars is C ( q ) = q 3 6 + 492 q + 400 , where q tons of steel are produced. The current level of production is 4 tons. Use calculus to estimate the amount by which manufacturer should decrease production to reduce the cost by 100 dollars. Solution. If 1 C denotes the change in C caused by a change 1 q in q , then for “small” values of 1 q we have, approximately, 1 C 1 q ≈ dC dq = q 2 2 + 492 . When q = 4, therefore, we have 1 C 1 q ≈ 500 , or (1) 1 C ≈ 500 1 q . We want 1 C to be 100; solving (1) for 1 q , therefore we want 1 q =  1 / 5. That is, we should decrease production by 1 / 5 ton. Of course, as the ≈ signs should signal to you, this is only ap proximate. The exact value could be solved by computing the cost when q = 4, C ( 4 ) = 7136 / 3, decreasing it by $100, C ( 4 ) 100 = 6836 / 3, and then solving q 3 6 + 492 q + 400 = 6836 3 . Easier said than done!! This requires solving a cubic equation; to give you an idea of how hard this is, the exact solution is q =  492 3 s 2 1409 + 5 √ 2461321 + 2 2 / 3 3 q 1409 + 5 √ 2461321 . Numerically, this is q ≈ 3 . 79984 , which means that we should decrease production by 0 . 20016 tons. This is one of the important uses of the derivative: it allows a quick estimate of the requisite quantity, as opposed to a long and difficult exact calculation. In this case, the estimate is particu larly good (particularly close to the exact value). Problem 2. Consider the curve defined by the equation 2 x 3 y y 3 + 5 x 2 y + 3 = . Find the equation of the tangent line to the curve at the point ( x , y ) = ( 1 , 2 ) . Solution. The equation of the tangent line is given by y 2 = m ( x 1 ), where m is the numerical value of dy / dx at this point on the curve. To compute dy / dx , we implicitly differentiate, d dx ( 2 x 3 y y 3 + 5 x 2 y + 3 ) = d dx , i.e., 6 x 2 y + 2 x 3 dy dx 3 y 2 dy dx + 5 2 dy dx = . Solving this last for dy / dx yields dy dx = 5 6 x 2 y 2 x 3 3 y 2 2 . That’s the general value of dy / dx ; we’re interested in the value when x = 1 and y = 2, m = 17 12 = 17 12 . Therefore the equation of the tangent line is y 2 = 17 12 ( x 1 ) , or y = 17 12 x + 7 12 . As a check, when we substitute x = 1 in this last, we do indeed get y = 2. While it isn’t necessary to solve the problem, a graph is shown in Figure 1.21.510.5 0.5 1 1.5 x1 1 2 3 y F igure 1. Implicit plot and tangent line at ( x , y ) = ( 1 , 2 ) . Problem 3. Let f ( x ) = 3 x 5 5 x 3 . Find: (a) The xcoordinate(s) of the critical point(s) of f ( x ) ....
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 Fall '07
 Vorel
 Calculus, Critical Point, dx

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