Find the absolute extrema

Question

Find the absolute extrema of f(x,y)= (x^2)y-2x+y on the closed triangle T with vertices (1,0),(5,0) and (1,4)

Answer 1

f(x,y)=x²y-2x+y

Notice that the partial derivative of f with respect to y is f_y=x²+1. Since:

f_y>0 -> x²+1>0 -> x²>-1 -> x is any real number

it follows that f(x,y) increases with increasing values of y for all values of x. Therefore, the largest value of f must be found somewhere on the upper diagonal edge of our triangular domain, which is the line segment from the point (5,0) to (1,4). The equation for the line passing between those two points can be found using the point-slope formula:

(y-0)/(x-5)=(4-0)/(1-5) -> y/(x-5)=-1 -> y=-(x-5) -> y=-x+5

If we plug in this expression for y into our equation for f(x,y) we obtain the values of f along this line segment as a function of x:

f(x,-x+5)=x²(-x+5)-2x+(-x+5)=-x³+5x²-3x+5

To find the local extremum of this function in the domain 1≤x≤5 we can take its derivative with respect to x and set it equal to 0:

f'(x,-x+5)=-3x²+10x-3=-(x-3)(3x-1)=0 -> x=1/3, 3

Since x=1/3 lies outside our domain it follows that the local extremum of our function must be located at x=3. The corresponding y-value is y=-3+5=2. Plugging in the point (3,2) into f(x,y) we obtain:

f(3,2)=3²*2-2*3+2=14

To determine if this point is a local maximum or minimum we need only check the value of f(x,y) at one of the two endpoints of the line segment. If we choose (5,0) to be this endpoint we obtain:

f(5,0)=5²*0-2*5+0=-10

Since 14>-10 it follows that (3,2) is a local maximum, and thus f(x,y) has a maximum extremum on the given triangular domain at the point (3,2) whose value is 14.

To find the minimum extremum of f(x,y) on the given triangular domain we simply note that since f_y>0 for all values of x it follows that this minimum extremum must be located somewhere on the bottom horizontal edge of the triangular domain. Since the equation of this edge is y=0 it follows that the value of f along this edge must be:

f(x,0)=x²*0-2x+0=-2x

Since -2x is a strictly decreasing linear function of x it immediately follows that on the domain 1≤x≤5 f(x,0) is minimized at x=5. Thus the point (5,0) is the minimum extremum of f(x,y) on the given triangular domain, and the value of f(x,y) at this point is -10 (as we have already seen).

Summary:

Maximum Extremum: f(3,2)=14

Minimum Extremum: f(5,0)=-10