11. What is the minimum value of f(x,y) = x2y for {(x,y): 4x2 + y2 = 4}?

A. –2/31/2

B. –4/3(31/2)

C. 0

D. 4/3(31/2)

E. 2/31/2

 

Solution: The answer is B

 

Since, 4x2 + y2 = 4, solve for x2 in terms of y and substitute into f(x,y), as follows:

To find the minimum value of f(x,y), first find the critical points of equation (1) and then test the corresponding intervals to determine the minimum value.

The critical points are:

The following graph illustrates the test of the corresponding intervals:

Which shows that the derivative goes from a negative value to a positive value at y = –2/31/2. Thus, substituting this value into equation (1) above, results in the minimum value of f(x,y).