Math 10C Practice Final Solutions

Transcription

1 Math 1C Practice Final Solutions March 9, (6 points) Let f(x, y) x 3 y + 12x 2 8y. (a) Find all critical points of f. SOLUTION: f x 3x 2 y + 24x 3x(xy + 8) x,xy 8 y 8 x f y x 3 8 x 3 8 x So even though x works for f x we need x 2 to have f y. So then y 8/2 4 ANSWER: (2, 4) (b) Classify each critical point of f as a local maximum, local minimum, or saddle point. SOLUTION: We use the D-test for functions of two variables. D f xx f yy f 2 xy (6xy + 24)() ( 3x 2) 2 9x 4 If you nd typos (which is likely) or errors in logic (hopefully less likely) in these solutions, please let me know either on Piazza or by (jbonthiu@ucsd.edu) 1

2 We can already see that, since x 4 > for all x, then clearly D < for all x, without even plugging in our critical point. Hence, any critical point is a saddle. ANSWER: (2, 4) is a saddle point 2. (6 points) Find an equation for the plane passing through the points A (1, 1, 5), B (1, 3, 1), and C (6, 1, 1). SOLUTION: We rst need to nd a vector that will be normal to the plane passing through all three points. We can do this by nding the cross-product of two vectors in the plane. Let's choose AB and AC (any vectors will do) AB i 4 j 4 k AC 5 i + j 4 k Then a normal vector to the plane will be given by AB AC i j k i(16 ) j( ( 2)) + k( + 2) 16 i 2 j + 2 k So one normal vector will be given by 16 i 2 j + 2 k. Let's scale this vector down by dividing by 4, since this looks kinda ugly. Thus, n 4 i 5 j + 5 k. Now we can use the equation for a plane passing through a given point with a given vector (choose any of the above points I choose A) 4(x 1) 5(y 1) + 5(z 5) 4x 5y + 5z x 5y + 5z 24 Note: your nal answer might look dierent than this. To check if it's correct, test all three points to see if they lie on the plane, and check if the normal vector is perpendiculr to at least two of the vectors in the plane (e.g. AB and AC) ANSWER: 4x 5y + 5z (6 points) Let v i + j k and w 2 i j + k. 2

3 (a) Compute v w. SOLUTION: We just compute it. v w ( 1) + ( 1) ANSWER: (b) Compute v w. SOLUTION: We just compute it. v w i j k i 3 j + 3 k 3 j 3 k i(1 1) j(1 ( 2)) + k( 1 2) ANSWER: 3 j 3 k (c) Find the angle between v and w. You may express this angle as the inverse cosine (or arccosine) of a number. SOLUTION: Part (a) helps two vectors have dot product of zero if and only if they are perpendicular. So... ANSWER: π/2 4. (6 points) A rectangular box without a top has a volume of 32 cm 3. Find the dimensions of the box having minimal surface area. SOLUTION: Let's set up the equations rst. We have the volume of the box, V, given by V (l, w, h) lwh, and the surface area of the box without the top, S(l, w, h) wl + 2wh + 2lh. We also know that V 32; that's xed (our constraint). Thus, our constraint is lwh 32,or h 32. We take this and plug it into the function for surface wl area to get a function of two variables, w and l: Now we minimize this function: S(w, l) wl + 2w wl + 64 l ( ) l wl + 64 w S w l 64 w 2 l 64 w 2 3 ( ) 32 wl

4 S l w 64 l 2 w 64 l 2 From this, we see that w l. Then we substitute w l in the rst equation: l 64 l 2 l 3 64 l 4 w 4 h Thus, (w, l, h) (4, 4, 2) is a critical point. minimum. We need to make sure that this is a S ww (4, 4, 2) 128 w 3 2 S ll (4, 4, 2) 128 l 3 2 S lw (4, 4, 2) 1 D(4, 4, 2) > S ww > local min So (4, 4, 2) really does minimize the surface area subject to the constraint. ANSWER: (4, 4, 2) 5. (6 points) Find the maximum and minimum values of the function f(x, y) xy subject to the constraint 9x 2 + y SOLUTION: We use Lagrange multipliers. Let g(x, y) 9x 2 + y 2 gradf λgradg f x i + f y j λ(g x i + g y j) y i + x j λ(18x i + 2y j) 4

5 We can equate each i and j component individually: y 18λx x 2λy Then by pluggin the bottom equation into the top, we get y 18λ(2λy) 36λ 2 y y, or 1 36λ 2 λ ±1/6 Then pluggin this into our equation y 18λx, we get y ±3x. Then plugging this into our constraint, we get: 18 9x 2 + y 2 9x 2 + (±3x) 2 9x 2 + 9x 2 18x 2 1 x 2 x ±1 Be careful here. Remember that we have two λ s: λ 1/6, and λ 1/6. λ 1/6 : y 3x x 1 : y 3 (1, 3) x 1 : y 3 ( 1, 3) λ 1/6 : y 3x x 1 : y 3 (1, 3) x 1 : y 3 ( 1, 3) So we have four critical points here: (1, 3), ( 1, 3), (1, 3), ( 1, 3). From here, we can just test the values of the function and nd out what maximum and minimum values it attains. But just for fun, let's classify each point. To classify these, let's take a look at the contour diagram: 5

6 By considering the points where the constraint (in blue) is tangent to contours of the function (in pink) we can view whether each contour is a maximum or a minimum. Thus, (1, 3) and ( 1, 3) are local maximums, and ( 1, 3) and (1, 3) are local minimums. (Note: we see also that we can just test the value of the function at each point, but I wanted to emphasize this method in case you need to use it). ANSWER: maximum at 3, minimum at 3 6. (6 points) Let f be a function that has the contour diagram given below. 6

7 (a) Find the coordinates (to the nearest.2) of the local maxima and local minima of the function f. SOLUTION: So we need to nd points on this graph that are locally maximal and points that are locally minimal. Let's start by nding the critical points. It looks like (1, 3.2), (.8, 1.8), (1.8,.8), (2.1, 2.1), and (3.2, 1) are all critical points. It's pretty evident that (2.1, 2.1) is the location of a saddle-point, since two contours are crossing through each other, and also by examining the surrounding values of f. By examining the surrounding values, we see that (1, 3.2) and (1.8,.8) are local minimums, and (.8, 1.8) and (3.2, 1) are local maximums. ANSWER: (1,3.2), (1.8,.8) local min's, & (.8,1.8), (3.2,1) local max's (b) The function f also has at least one saddle point. Find the coordinates (to the nearest.2) of the saddle point(s). SOLUTION: Hey, look at that! We already did this in part (a). ANSWER: (2.1, 2.1) 7. (6 points) Evaluate the double integral R (x2 + xy)da, where R is the triangle in the xy-plane having vertices at (, ), (1, 1), and (1, ). SOLUTION: Let's sketch out the region that we're integrating over: 7

8 You can choose which direction to integrate over rst, but I'm going to do both, just 'cuz. Notice one direction is a little bit cleaner. y (x 2 + xy)dxdy ( ) x x2 2 y 1 y dy ( (1 y 3 ) + (1 y2 ) 2 3 ( 1 3 y 1 12 y y2 1 8 y4 ) y ) 1 dy ˆ x (x 2 + xy)dydx (x 2 y + xy2 2 (x 3 + x3 2 ( ) x x4 1 8 ) x ) dx Either way, we arrive at the same answer. 3 ANSWER: 8 8. (7 points) Evaluate the integral by reversing the order of integration 1 1 4ysin(x 2 )dxdy. y SOLUTION: 2 We can't integrate this thing as it's written you can give it a shot, 8

9 but it won't work out well. So we need to change the order of integration. Let's begin by sketching out our region of integration: So then after changing the order of integration, we have: ˆ x 4ysin(x 2 )dydx 2y 2 sin(x 2 x ) dx 2xsin(x 2 )dx Let u x 2. Then du 2xdx, and 2xsin(x 2 )dx sin(u)du cos(u) 1 cos(1) + cos() 1 cos(1) ANSWER: 1 cos(1) 9