Find the absolute extrema?

Question

Find the absolute extrema of f(x, y)=3x^2+2y^2-4y on the region in the xy-plane bounded by y=x^2 and y=4.

df/dx=6x

df/dy=4y-4

6x=0, 4y-4=0,

(0, 1)

g(x)=f(x, x^2)=3x^2+2x^4-4x^2=2x^4-x^2

dg/dx=8x^3-2x=0

x=0, 1/2, -1/2

What to do next?

Answer 1

Hi Sun,

That's a great start and you have almost solved the problem.  

You found the critical point at (0,1), but you need to find the discriminant in order to determine if this is the max or min of the function.  The discriminant is found as f_xxf_yy -f_xy² = 6*4 = 24.  Since both the discriminant and f_xx are positive, this is a minimum.  And since it is the only critical point of the function, it is the absolute minimum.  (It is also important to verify that this point is within the region, which it is). 

f(0,1) = -2

Now you need to find the critical points on the bounding curves y = x^2 and y = 4, as well as the points of intersection, and figure out which one gives the highest value - this will be the absolute maximum.  

Plugging x = 0, 1/2, -1/2 into g(x) gives

g(0) = 0; g(1/2) = -1/8; g(-1/2) = -1/8;

 

Similarly, we can say h(x) = f(x,4) = 3x² + 16;

f'(x) = 6x; >>> x = 0; y = 16

Finally, test the points of intersection (2,4) and (-2,4).  

f(2,4) = f(-2,4) = 28.  

Thus, the absolute max is at (+/-2,4) = 28 and the absolute min is at (0,1) = -2.