Math 2263 Fall 2013 - Quiz 4
10/9/2013
Name:
Please answer the following questions completely and clearly. An unsupported answer is worth few points.
(1) Find three positive numbers whose sum is 100 and whose product is maximum.
Solution:
We are trying to maximize f = xyz where x + y + z = 100. We need to write f as a function of two
variables using our constraint:
z = 100 − x − y =⇒ f (x, y) = xy(100 − x − y)
Taking the first partials and setting them equal to zero:
fx = 100y − 2xy − y 2 = 0, fy = 100x − x2 − 2xy = 0
Take first, solve for x to get x = 50 − y/2 and plug this into fy :
fy = 100(50 − y/2) − (50 − y/2)2 − 2y(50 − y/2) = (50 − y/2)(100 − (50 + y/2) − 2y) = 0
=⇒ y = 100, y = 100/3
If y = 100/3, then x = z = 100/3 also. If y = 100, we necessarily have that x = z = 0, which gives
f (0, 0, 0) = 0, and this won’t be a maximum. Let’s assume that we didn’t realize this, and test both points
in D = (−2x)(−2y) − (100 − 2x − 2y)2
D(0, 0) = −1002 < 0
And so (0, 0) cannot be where the max occurs.
D(100/3, 100/3) = 4(100/3)2 − (100 − 4(100/3))2 > 0 =⇒ fxx (100/3, 100/3) = −2(100/3) < 0
And so, the max occurs where x = 100/3, y = 100/3, and z = 100/3
(2) Use Lagrange multipliers to find the maximum and minimum values of the function f (x, y) = 3x + y subject
to the constraint x2 + y 2 = 10.
Solution:
Taking appropriate partials, we have that
3 = λ(2x), 1 = λ(2y)
Solving for λ in both and setting them equal, we get
3
1
=
=⇒ x = 3y
2x
2y
Plugging this into our constraint and solving:
(3y)2 + y 2 = 10 =⇒ y = ±1
1
Since x = 3y, we get two points: (−3, −1) and (3, 1). Plugging these into f , we have that f (−3, −1) = −10
is the minimum value and f (3, 1) = 10 is the maximum value.