Math 265 (Butler) Practice Midterm II B (Solutions)

Transcription

1 Math 265 (Butler) Practice Midterm II B (Solutions) 1. Find (x 0, y 0 ) so that the plane tangent to the surface z f(x, y) x 2 + 3xy y 2 at ( x 0, y 0, f(x 0, y 0 ) ) is parallel to the plane 16x 2y 2z 23. Rearranging we have F (x, y, z) x 2 + 3xy y 2 z 0 with gradient F 2x + 3y, 3x 2y, 1. In order for the tangent plane to be parallel to the plane 16x 2y 2z 23 we need to have that the normal vectors are parallel (i.e., scalar multiples of one another). So we need 2x + 3y, 3x 2y, 1 k 16, 2, 2. Comparing the last entries we need to have k 1/2 and so we need to solve 2x + 3y 8 and 3x 2y 1. Solving for x in the first equation we have x 4 (3/2)y which on substituting in the second equation gives 3(4 (3/2)y) 2y 1 or 13/2y 13 or y 2, showing that x 1. Therefore the desired point is (1, 2).

2 2. The Nook corporation (maker of the finest Yops) has recently merged with the Jibboo conglomerate (maker of the finest Zans). Currently Yops sell for three dollars each and Zans sell for nine dollars each. By combining their production the new company enjoys economy of scope and is now able to produce y Yops and z Zans at a cost of y z3 yz dollars. Determine how many Yops and Zans respectively should be made in order to maximize profit. Also, verify that your answer is a maximum by using the second derivative test. (Economic flashback: profit is the difference between the revenue you get from selling the products and the cost it took to produce the products.) Following the convention given we will let y be the number of Yops and z the number of Zans. Then the revenue will be 3y + 9z (i.e., three dollars for each of the Yops and nine dollars for each of the Zans). By our economic flashback the profit is the difference between revenue and cost, i.e., P (y, z) (3y + 9z) ( } {{ } 2 y z3 yz) 1 } {{ } 2 y2 1 3 z3 +yz+3y+9z 10. Revenue Cost To find the critical point, we set the partial derivatives to 0, i.e., P (y, z) y y + z P (y, z) z2 + y From the first equation we get z y 3, which if we substitute into the second equation becomes 0 (y 3) 2 + y + 9 (y 2 6y + 9) + y + 9 y 2 + 7y y(y 7). This gives us either y 0 or y 7. We can throw out the y 0 solution since then z 3 (and you can t produce negative Zans!). That leaves us with y 7 and z 4. Since this is the only critical point this must be the correct production level, so we should produce 7 Yops and 4 Zans. To verify that this is indeed a maximum (it would be embarrassing to set the company production levels to make the minimum profit!) we apply the second derivative test. We have P yy P zz (P yz ) 2 ( 1)( 2z) (1) 2 2z 1. Since at y 7 and z 4 we have that this is positive (i.e. 7) we are at a max or a min, and since P yy 1 < 0 we can conclude that this is indeed at a maximum. We never actually computed the profit. It is entirely possible that even when maximizing profit the company loses money. Fortunately in our case we will be making a nice profit of 175/6 (but this does not need to be computed to answer the question!) On a side note: the names of the products and companies are taken from Dr. Seuss books.

3 3. Sketch in the xy-plane the domain of f(x, y) 4 y 2 ln(y x 2 ). We look for our three potential problems, and in this case we have all sorts of problems. Division by zero: The denominator will be zero when the term inside of the log is 1, therefore our domain has the restriction that it is all (x, y) so that y x 2 1 or y x Square root of a negative: We need that the term inside the square root is non-negative, therefore our domain has the restriction that it is all (x, y) so that 4 y 2 0, or y 2 4 which is equivalent to 2 y 2. Log of a non-positive: We need that the term inside the log is positive, therefore our domain has the restriction that it is all (x, y) so that y x 2 > 0, or y > x 2. Putting all of these together we get the following picture. y x y x 2 y 2

4 4. Given the implicitly defined surface z + sin z xy find 2 z/ y only in terms of z. (Hint: find both / and / y then take the derivative of one of these with respect to the appropriate variable to find 2 z/ y, at the end a substitution from the original relationship defining the surface then gives the desired form.) Let us follow the hint. First we compute the two partial derivatives. Taking the derivative of both sides with respect to x and with respect to y we have the following (remember we think of z z(x, y)): + (cos z) y y x or y + (cos z) y or We now use the quotient rule to find 2 z y ( ) ( y x 1 + cos z ) y 1 + cos z, x 1 + cos z. (1 + cos z) x( sin z) (1 + cos z) 2 ( ) y (1 + cos z) + x sin z 1 + cos z (1 + cos z)2 + xy sin z (1 + cos z) 2 (1 + cos z) 3 (1 + cos z)2 + (z + sin z) sin z (1 + cos z) 3. At the last step we used the original relationship z + sin z xy to replace the xy term. You can also simplify it further using the Pythagorean identity to get the following (though the form given above is perfectly fine): 2 z y cos z + z sin z (1 + cos z) 3.

5 5. Show that u(x, t) sin ( x + sin t ) is a solution to the partial differential equation u t u xx u x u tx. To see if it is a solution we need to compute the various partial derivatives and see if this function satisfies the equation. We have the following: u x cos ( x + sin t ) Therefore we have u t cos ( x + sin t ) cos t u xx sin ( x + sin t ) u tx sin ( x + sin t ) cos t u t u xx ( cos ( x + sin t ) cos t )( sin ( x + sin t )) ( cos ( x + sin t ))( sin ( x + sin t ) cos t ) u x u tx showing that u is a solution to the differential equation.