If f(x , y) = x^2+3y-3xy a) Find the local minimum value, maximum value and sadd

Question

If f(x , y) = x^2+3y-3xy a) Find the local minimum value, maximum value and saddle points. b) Find the absolute minimum and maximum of the function f on the region bounded by y = x, y = 0 and x = 2 c) Find absolute minimum and maximum of the function f bounded by the region x^2+y^2 < or = 16

Explanation / Answer

F(x,y) = 3xy-x²y-xy² F? = 3y - 2xy - y² F? = 3x - x² - 2xy To find the stationary points we set these to zero and solve the equations for x & y 3y - 2xy - y² = 0 … (i) 3x - x² - 2xy = 0 … (ii) To solve first we need to find one variable in terms of the other. (i) - (ii) gives 3y - 3x - y² + x² = 0 ===> 3(y-x) - (y-x)(y+x) = 0 ===> (3-y-x)(y-x) = 0 ===> either y+x = 3 or y=x If y=3-x then (ii) becomes 3x-x²-2x(3-x) = 0 ===> x(x-3) = 0 ===> x=0, 3 So two points are ( 0, 3 ) and ( 3, 0 ) If y=x then (ii) becomes 3x-x²-2x² = 0 ===> 3x(1-x) = 0 ===> x=0, 1 So two points are ( 0, 0 ) and ( 1, 1 ) To find the type of stationary point first we find the partial second derivatives F?? = -2y, F?? = -2x, F?? = 3-2x and compute det(H) = F??F??-F??² For each stationary point, we examine these values ( 0, 3 ) : F?? = -6, F?? = 0, F?? = 3, det(H) = -9 Since det(H)

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