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## Calc Bus & Soc Sci II

by: Henderson Lind II

10

0

5

# Calc Bus & Soc Sci II MATH 242

Henderson Lind II
UO
GPA 3.8

Staff

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COURSE
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Staff
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PAGES
5
WORDS
KARMA
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## Popular in Mathematics (M)

This 5 page Class Notes was uploaded by Henderson Lind II on Tuesday September 8, 2015. The Class Notes belongs to MATH 242 at University of Oregon taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/187189/math-242-university-of-oregon in Mathematics (M) at University of Oregon.

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Date Created: 09/08/15
Business Calculus H Final Review Sheet Remember our nal is on TUESDAY MARCH 15 at 315pm 1 to CA3 Calculate the following inde nite integrals a fez 1 z Udz 2 3 f f idr 2 C l d 312 2e g2 dz 2 e f 43 2ln13 21dz Solution a x2xlnlxla b Let u x3 216 Then if 3x2 2 S0 3x2 2dc du Hence the integral transforms to f n c ln lxa QIl c 0 Let u x3 2x again so 332 2dc du The integral transforms to 7 8 72acg2ac2 C d Let u x3 2x again so 3x2 2dc du The integral transforms to feudu e c ewg2 c e Let u lnx3 2x so if 373 30 du di The integral transforms to fudu 72 c w c Calculate the following de nite integrals a fol zgdz 1 2 b f0 Tarldz c 01 zexzdz Solution 4 6 lel5 1 b Let u x3 1 Then du 3x2dx S0 xzdx du The integral transforms to ff nu an 0 Let u x2 Then du Qxdx s0 xdx du The integral transforms to H 6 ieui ace 1 Calculate the areas of the following regions Do not count areas below the zams as negative a The region enclosed by y z and y 714 b The region between y z y 13 z 71 and z 1 c The region enclose by y e y 2 and the y axis Solution F 533 a Where do they cross x 7x4 so x 0 O or x 71y 71 So we need to compute f317c4 7 xdx You should sketch the graph to see what is going on This is 7x557x22O71 71571212715 5107210 310 bWheredoycandyc3 cross xx3 ifx00rx1orx71 If you draw the picture you see that this region is made of two equal areas So the total area is twice the right hand area7 ie 2 01x 7 x3dc 2x22 7 5145 212 71412 c Again you need to draw a sketch If you do you see that you can compute this area by subtracting folquot 2 exdx from the area of the rectangle with base 1n and height 2 So the area is 21n2 7e 13 2 21n2 7 2 1 anQ 71 Sales of ice cream are continuously rising at a rate of 10 percent per month My ice cream company currently sells 1000 quarts each month Write down a differential equation describing the change in sales and then solve it to predict my monthly sales in 6 months time Solution Let 5t be monthly sales after t months The initial condition is that 50 1000 The differential equation is s t O1st So 015 Separate Variables to get I d7 01fdt so lns 01t 0 Apply 67 to both sides to get 5 1460 it for some new constant A ea Plug in t O to get 1000 A Hence 5 100060 it So in 6 months time the monthly sales will be 5 1000cO 6 1822 quarts Find the equation for the tangent plane to the following surfaces at the point 17 2 3 7 2 7 Z 7 a 271 31y y 3 b 2 7x73y2 Solution a 2x3y g7 3x At x 1 and y 2 these giVe8 and 35 So the tangent plane is z 8x 35y7 To nd the 7 put x 1 y 2z 3 to get 3 157 so 7 712 So its z 8x 35 712 b Duh lts already a plane So there is nothing to do The answer is z 7x 7 3y 2 Find all the critical points of the following function7 and decide if they are local minima7 loca maxima or neither a ay 12y 7 212 7 4y b ay 12 y7 ey 7 2 2 2 C ay 7r y Jr a my Solution a fag Qxy 7 4x O7 fy x2 7 8y 0 We need to solVe these simultaneous equations From the rst one7 we have x2y 7 4 0 so either 7 x 0 or y 2 If x 0 the second one gives y 0 too If y 2 the second one gives x2 16 so x i4 So there are three critical points 00 4 2 74 2 To decide What they are let us use the second derivative test We need fyyy 7 y That is H 21747872x2 32 7161 74x2 Also 2y 74 At 00 that gives H 32 and 74 So 00 is a local maximum At i42 that gives H 0 So the second derivative test doesn t help at all here So noW We have to think about some slices First take the slice x i4 The function then is 16y 7 32 7 412 This is an upsidedoWn parabola With vertex at y 0 its a maximum on this slice Consider instead the slice When x i2y The function then is 4y3 7 12y2 and We re interested in the place Where y 2 This is a cubic With a local minimum at y 2 So its a local minimum on this slice So its neither a max or a min overall b fan 2x and fy 1 7 ey so the only critical points is at 00 When these both vanish Also famfyyy 7 72 and 2 at the point 00 So its a local minimum c fag 2x 7 57y 0 fy 2y 7 x72 0 Solving these simultaneous equations you get xay 1 and xya 1 The rst says that y 1163 the second gives xxg 1 so x x9 doesn t solve x y 1 so this one doesn t count So either x 1 When y 1 or x 71 When y 71 So there are tWo critical points 11 and 7171 Substituting into soeitherx00rci1 Butxi Now fxyxfyyy 7 at either of these points Works out to 32 and Works out to 6 So both points are local minima d fag 2xexzy2 0 fy 2yexzy2 0 So critical points just at 00 Thinking about 673 the exponent is always 2 0 so the smallest 673 can ever be is co Which happens at the origin So its a global minimum On my farm in Portugal I sell cabbages and asparagus at prices p1 and p2 Euros respectively The demand function for cabbages is given by 11 23 i 1001 102 The demand function for asparagus is given by 12 172101 802 What prices should 1 charge in order to maximize revenue Solution R 171 quz 17123 i 10171 172 p217 21717 8172 23171 7 10p 17172 7 817 3171172 Find the critical points 23 7 20171 3172 0 17 7 16172 3171 0 To solve this system of linear equations put it into the augmented matrix 20 73 1 The reduced roW echelon form is 1 0 319311 0 1 409311 73 23 16 17 00 H O Hence we should take p1 319311 and p2 409311 The Portuguese mail service will accept only packages of length no more than 270 cm and length plus girth no more than 325 cm What are the dimensions of the largest package that 1 would be able to mail back to Eugene from my farm in Portuga Solution Let the length be 2 the width be y and the height be 2 We need to maximize V xyz Volume The constraint is that 2 2y 22 325 Use this to eliminate x we have V 325 7 2y 722yz 325yz7 21227 2122 Compute critical points 3252 7 4yz 7 222 O7 325g 7 2y2 7 4yz O The solution is certainly not going to be when 2 0 or y 0 So we have a right to divide the rst equation by z and the second equation by y Then we need to solve 32574y72z0 32572y74z0 2 325 4 2 4 325 The rref is You can solve this using matrices The matrix is 1 0 3256 7 7 7 0 1 3256 Sowemusttakeyizi3256Thenxi325 2y 227 3253 The Volume is 34328125108 Find the general solution to the following system of linear equations 21 3y 62 0 4x 5y 62 3 7x 8y 92 6 Solution You put the matrix 2 3 6 0 4 5 6 3 7 8 9 6 in to your calculator then rref it to get Hence the solution is xy2 327 07 712 If I and y are two numbers summing to 8 what is the biggest possible value their product my could be Solution We have that x y 8 We want to maximize p 2y Eliminate y Weneed to maximizepx87c827x2 872xOwhenx4 So the maximum occurs when x y 47 and their product is 16 11 Solve the following differential equation subject to the initial condition that y 0 When I 0 dy 7 1 1 d1 Iy Solution Sepamate Variables to get f Tyl fxdx S0 111W 1 I22 c Set x y O to get 0 0 S0 111W 1 I22 D0 67 to both sides to get y 1 67 22 S0 y 67 22 7113 the solution

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