I need complete solutions Respond to each of the following prompts to solve the

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I need complete solutions

Respond to each of the following prompts to solve the given optimization problem. Let f(x, y) = sin(x) + cos(y). Determine the absolute maximum and minimum values of f. At what points do these extreme values occur? For a certain differentiable function F of two variables x and y, its partial derivatives are F_x(x, y) = x^2 - y - 4 and F_y(x, y) = -x + y - 2. Find each of the critical points of F, and classify each as a local maximum, local minimum, or a saddle point. Determine all critical points of T(x, y) = 48 + 3xy - x^2y - xy^2 and classify each as a local maximum, local minimum, or saddle point. Find and classify all critical points of g(x, y) = x^2/2 + 3y^3 + 9y^2 - 3xy + 9y - 9x

Explanation / Answer

(a) f(x,y) = sin(x) + cos(y)

We know that -1<=sinx<=1 and also -1<=cos(y)<=1

Hence -1-1<=sin(x)+cos(y)<=1+1

=> -2<=sin(x)+cos(y)<=2

Hence absolute minimum value of f = -2 and absolute maximum value is 2

(b) Fx = 0 => x2 - y - 4 = 0

Fy = 0 => -x + y - 2 = 0 => x-y = 2 => x-2= y

Hence x2-x+2-4 = 0 => x2-x-2=0 => x(x-2)+1(x-2)=0 => x=-1, x=2

Then y = -1-2 = -3, and y = 2-2 = 0

Hence critical points are (-1,-3) and (2,0)

Now consider D = fxxfyy-(fxy)2

Now fxx = 2x, fyy = 1, fxy = 0

Hence D = 2x

For (-1,-3); D = -2<0, hence (-1,-3) is saddle point

For (2,0); D = 4>0, then see fxx = 4>0 Hence (2,0) is point of minimum of f(x,y)

(c) Tx = 3y-2xy-y2 = 0 => y(3-2x-y) = 0 => y=0 or 3=2x+y

Ty = 3x-x2-2xy = 0 => x(3-x-2y) = 0 => x=0 or 3=x+2y

Hence we get x=0,y=0 or y=0,x=3 or x=0,y=3 or x=1,y=1

Then critical points are (0,0), (3,0), (0,3) (1,1)

Now Txx = -2y, Tyy= -2x, Txy = 3-2x-2y

D = TxxTyy-(Txy)2 = 4xy-(3-2x-2y)2

At (0,0) ; D = -9<0Hence (0,0) is saddle point

At (3,0) ; D = -9<0 Hence (3,0) is saddle point

At (0,3); D = -9<0 Hence (0,3) is saddle point

At (1,1) ; D = 4-(3-2-2)2 = 4-1 = 3>0; Now Txx = -2<0 Hence (1,1) is absolute maximum point

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