Download MAT 16C: Quiz 3 - Practice

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MAT 16C: Quiz 3 - Practice
Monday, July 11
Write legibly and neatly. Show all your work for full credit.
1. (10 points) Use Lagrange multipliers to minimize the function:
f (x, y) = x2 − 8x + y2 − 12y + 48,
subject to the constraint:
x + y = 8.
Please provide both the point at which the function is minimum, as well as the minimum value the
function takes on. You will not receive any credit for solving this problem in a way that does not use
Lagrange multipliers. You do not need to justify the fact that the point you find is a minimum.
We begin by identifying F(x, y, λ ). We can rewrite the constraint equation as x + y − 8 = 0, and so:
F(x, y, λ ) = x2 − 8x + y2 − 12y + 48 − λ (x + y − 8)
Then we take partial derivatives:
Fx = 2x − 8 − λ
Fy = 2y − 12 − λ
Fλ = −x − y + 8
We can then set these all to 0 and solve for x and y. I’ll do this using the method I call ‘solve for everything
in terms of lambda’. I’ll start with the first equation:
0 = 2x − 8 − λ
2x = 8 + λ
x = 4 + λ /2
Solving for y in the second equation yields:
0 = 2y − 12 − λ
y = 6 + λ /2
We then plug into the final equation:
0 = −x − y + 8
8 = 4 + λ /2 + 6 + λ /2
λ = −2
Plugging back into the previous equations, we see that x = 4 + λ /2 = 3, and that y = 6 + λ /2 = 5. So the
minimum occurs at (3, 5), and the minimum value of f is given by:
f (3, 5) = 32 − (8)(3) + 52 − (12)(5) + 48
= 9 − 24 + 25 − 60 + 48
= −2
2. (5 points) Evaluate the iterated integral:
Z 2 Z 2y−4y2
0
3y dx dy
3y2 −6y
We just compute:
Z 2 Z 2y−4y2
0
3y2 −6y
Z 2y−4y2
Z 2
3y dy dx =
3y
0
Z 2
=
0
3y2 −6y
dx dy

2
 2y−4y

 dy
3y x
2
3y −6y
Z 2
=
3y (2y − 4y2 ) − (3y2 − 6y) dy
0
Z 2
=
3y 8y − 7y2 dy
0
Z 2
Z 2
y2 dy − 21
y3 dy
0
0
 
 
2
2
3
y 
y4 


= 24
− 21
3
4
= 24
0
= 24(8/3) − 21(4)
= 64 − 84
= −20
0
3. (10 points) Let f (x) = x, and g(x) = x2 .
0.5
1.0
(a) Sketch the region bounded by these two functions. (Hint: sketch both functions on the same graph.
They intersect in two places, and form a contained area in the first quadrant. This is the region we are
interested in.)
−0.5
0.0
y
R
−0.5
0.0
0.5
1.0
x
(Note that you should shade in the region R... its just hard to make my computer do that!)
(b) Use a double integral to find the area of this region.
You could choose to do this in either direction. I will integrate in y first, and then in x, since the functions
are already both given in terms of y (not that its hard to change them).
Z 1Z 1
Z 1
1 − x2 dx
0
1
x3 = x− 3
dy dx =
0
x2
0
= 1 − 1/3
= 2/3