Expert: Scott A Wilson - 12/31/2008

Question
QUESTION: I have two questions, but i think they have some difference between them??

1. Find the maximum and minimum values of
f (x,y) = xy on the ellipse 9x^2+y^2 = 6.
maximum value =
minimum value =
2. ( Find the maximum and minimum values of
f (x,y) = 5x^2+6y^2 on the disk D: x^2+y^2 <= 1.
maximum value:
minimum value:

ANSWER: 1. There may be a shorter way to do it, but I think of xy as a product of two functions.  Note that y = ±√(6-9x²).

For both problems, solve the constraint for y and put into the first equation.  For the 2nd constraint, look into it to see where the minimum is at.  More mention of that a little later.

1. To find the maximum and minimum, take the derivative and set it equal to 0.  The problem is the product rule with f(x)=x and
g(x)=±√(6-9x²).  Note that f'(x)=1 and g'(x) = -±9x/√(6-9x²).  Note that the -± should really have the minus sign on top, but the computer can't do that.

The second derivative will tell you if they are a maximum or a minimum.  Note that computing a second derivative will probably take a few minutes.

By the product rule, we need to find fg' + gf', which turns out to be
-±9x²/√(6-9x²) ±√(6-9x²).  Setting this to 0 and multiplying the entire equation by the denominator gives ±9x² = ±(6-9x²). The ± are the same now since the 6-9x² was moved to the other side of the equation.  I will take them both as + right now.  Add 9x² to both sides and 18x² = 6 is gotten, so this can quickly be solved for x.


2. Let y = ±√(1-x²).  Put this in to f(x,y) so now you only have
f(x).  Take the derivative of f(x), set equal to zero, and look at the values.  Since the curve is circular, the minimum value can be picked out.  Note that f(x,y) is the sum of two independent parabolas, so the minimum of each parabola is at 0.  The only value that's left to find is the maximum.


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

QUESTION: Thank you for answering my question.But i think i am missing something. After finding the value for x (in 1st question) i have to Substitution the value of x in the function again to find y.And after doing that i will have one point (x,y).
In other words, from which function i should find the maximum and minimum, after doing the derivative.

ANSWER: For 1, I get the following:

 The x,y values are on an ellipse 9x² + y² = 6.
 The extreme points for the function are when x =0.577350269.  
 When y is 1.732050808, the function is at a maximum (xy=1).  
 When y is -1.732050808, the function is at a minimum (xy=-1).

For 2, I get the following:

 The extreme points for the x are when x=0.
 When y=1, the fuction is at a maximum.
 When y=-1, the function is at a minimum.

1) Now to answer the question.  When finding the extreme points of each function, put the value of x back into the ellipse to find the y value.  Once this has been found, put x and y into xy to find the functional max and min.

2) Now to answer to second question.  When finding the extreme values for this one, put the x value back into the circle to find the y value.  Put x and y back into 5x² + 6y² to get the answer.  Note that here, x is 0, so all that has to be done to find the extreme values is put in the value for y.  Since y is ±1, the function is ±6 at the extreme points.

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

QUESTION: Thank you again doctor, i really understand the your answers. And to make sure that i am understanding well i tried to solve another question that is similar to the first question, but the answer was not correct, so i post my question with answer to check it to me where is the wrong step exactly.
Q: Find the maximum and minimum values of f(x, y) = xy on the ellipse 5x^(2) + y^(2) = 9.
maximum value =
minimum value =

My Answer:
f`(x,y)=1  and g`(x)=+- 5x/(9-5x^2)
-+-(5(x^2))/(9-5x^2)+(9-5x^2)^1/2=0
10x^2=9  ,so x=+-3/(10)^1/2
then Substitute in the function:
5*(9/10)+y^2=9   then y=9-5*(9/10)=+-4.5
when y=4.5 the function(x,y)=4.27
when y=-4.5 the function (x,y)= -4.27


Answer
f'(x,y) is really 1 + dy/dx, since this is a product of x and y.

To do this correctly, y must be solved for in the ellipse and this put back into f(x,y) so that it becomes f(x).

Solving the ellipse for y gives us y=±√(9-5x²).

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).
This can be rearranged if the terms are combined by multiplying the second term by itself over itself so that they have a common denominator.

To see where the slope is undefined, it is where 9-5x²=0.
To see where the maximum(s) and minimum(s) occur, it is where
-5x² + √(9-5x²) = 0.  This can be changed to 5x² = √(9-5x²) and then both sides can be squared.  It is a quadratic equation with x² as the varialbe (since it has an x^4 and and x² in it).  Solve for x² with the quadratic formula, and then remember that x can be both positive and negative.

Once this has been done, put the values of x back in the equation to see if they are good and then determine if each is a maximum or a minimum.