Expert: Sherry Wallin - 1/3/2009

Question
QUESTION: Q: Find the maximum and minimum values of f(x, y) = xy on the ellipse 5x^(2) + y^(2) = 9.
maximum value =
minimum value =

Putting this back into f(x,y) = xy gives us
f(x) = ±x√(9-5x²).

This is a product of g(x) = x and h(x) = √(9-5x²).
The derivative of f is gh' + hg'.

f'(x) = (x) (1/2)(1/√(9-5x²))(-10x) + √(9-5x²)(1)
= -5x²/√(9-5x²) + √(9-5x²).
-5x² + √(9-5x²) = 0.  This can be changed to 5x² = √(9-5x²)
then squared the both side to find x^2 and then find x
but the problem her is  when i find x^2, the answer will be negative[ x^2=(-5*+5*sqrt37)], then i can not take the square root of x^2 to find x.
Could you see where is the wrong step, please?

ANSWER: Miss Adel~
    As I follow your process to solve for x I see nothing wrong until the step where you say "the problem here is when I find x^2, the answer will be negative[ x^2=(-5*+5*sqrt37)". Remember when you take the square root you get two possible answers and only one of them is negative. I get x^2 = [-1+-sqrt(37)]/10 and [-1+sqrt37]/10 is approx .508 which is a positive number so x is approximately .71. I hope this has cleared your problem for you.

Math Prof

You have [ x^2=(-5*+5*sqrt37) and have not divided by 10. Also it looks like you are multiplying 5*sqrt37 by negative 5 and the quadratic formula says [-b+-sqrt(b^2-4ac)]/2a. In other words plus minus not multiply.

---------- FOLLOW-UP ----------

QUESTION: Thanks very much. I still has one more question the value of x should be replaced in the function of ellipse and then from the value of x and y replaced in the 2nd function to find the Min and Max values? Please tell me if i am right or not?

Answer
Miss Adel~
    I think I have an easier way to solve this problem so I will try to explain to you step by step how to solve this problem. I will use rt( ) to mean the square root of ( ), so rt(9)= 3 etc

First find the partial derivative of f(x,y) with respect to x and y.
f_x(x,y) = y and f_y(x,y) = x

Make the equation of the ellipse your g(x,y)
Now set each partial of f equal to the corresponding partial of g allowing a multiple of each. What I mean by that is set
f_x(x,y) = kg_x(x,y)  and f_y(x,y) = kg_y(x,y) where k is some constant.

5x^(2) + y^(2) = 9 gives us 5x^(2) + y^(2) - 9 = 0
So f_x(x,y) = kg_x(x,y) gives us y = k(10x) and
f_y(x,y) = kg_y(x,y) gives us x = k(2y)

Now substitute back into f(x,y) for y:  x = k(2(k10x))= 20xk^2 and set it equal to 0: x-20xk^2 = 0, factor: x(1-20k^2) = 0 so either x = 0 or
1-20k^2 = 0-> 1 = 20k^2-> (1/20) = k^2-> k = +-rt(1/20) = +-rt(5)/10
so we have x = 0 or k = +-rt(5)/10, now plug back into g(x,y) and get:

when x = 0 then 5x^(2) + y^(2) - 9 = 0: y^2 = 9 and y = +- 3 so we have the points (0,3) and (0, -3) as possible extrema of f(x,y). Do the same for k = +-rt(5)/10 and find the points (3rt(10)/10,3rt(2)/2), (3rt(10)/10,-3rt(2)/2), (-3rt(10)/10,3rt(2)/2), (-3rt(10)/10,-3rt(2)/2).

Now substitute the six points you have found into f(x,y) to find the max and min values of f(x,y): 0
(x,y): (0,3),(0,-3),(3rt(10)/10, 3rt(2)/2),(-3rt(10)/10, 3rt(2)/2)
f(x,y):   0       0          9rt(5)/5          -9rt(5)/5

(x,y): (3rt(10)/10, -3rt(2)/2), (-3rt(10)/10, -3rt(2)/2)
f(x,y):   -9rt(5)/5          9rt(5)/5

You can see that f(x,y) takes on a maximum value of 9rt(5)/5 at both
(3rt(10)/10, 3rt(2)/2) and (-3rt(10)/10, -3rt(2)/2) and a minimum value of -9rt(5)/5 at(-3rt(10)/10, 3rt(2)/2)&(3rt(10)/10, -3rt(2)/2).

I am available to explain in further detail if you need me to.

Math Prof