Global Extremum – Domain of ellipse – Exercise 4749

Exercise

Find the maximum value and the minimum value of the function

z(x,y)=x2yz(x,y)=x-2y

In the domain:

D={(x,y):x24+y22}D=\{ (x,y): \frac{x^2}{4}+y^2\leq 2\}

Final Answer


maxDz(2,1)=4,minDz(2,1)=4\max_D z(2,-1) = 4,\min_D z(-2,1) =-4

Solution

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