solve please Consider the function f(x, y) = e4x-x2-8y-y2. Find and classify all

Question

solve please

Consider the function f(x, y) = e4x-x2-8y-y2. Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank. fx = fy = fxx = fxy = fyy = The critical point with the smallest x-coordinate is ( , ) Classification: (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is ( , ) Classification: (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is ( , ) Classification: (local minimum, local maximum, saddle point, cannot be determined)

Explanation / Answer

fx = e^(4x-x^2-8y-y^2) * (4-2x)

fy = e^(4x-x^2-8y-y^2) * (-8-2y)

fxx = e^(4x-x^2-8y-y^2)* (-2) + (4-2x)^2 * e^(4x-x^2-8y-y^2)

fyy = e^(4x-x^2-8y-y^2)* (-2) + e^(4x-x^2-8y-y^2) * (-8-2y)^2

fxy = (4-2x) * e^(4x-x^2-8y-y^2) * (-8-2y)

so find the critical points we equate fx=0 and fy=0

since e^(4x-x^2-8y-y^2) can never be zero , from fx=0 we have 4-2x=0 so x=2

since e^(4x-x^2-8y-y^2) can never be zero , from fy=0 we have -8-2y=0 so y=-4

SO ONLY CRITICAL POINT = (2,-4)

fxx at (2,-4) = e^(20) * (-2+0) = -2e^(20)

fyy at (2,-4) = e^(20) * (-2) = -2e^(20)

fxy at (2,-4) = e^(20) * 0 = 0

so fxx * fyy - (fxy)^2= 4e^(40) which is >0 and here we can see that fxx<0

so fxx * fyy - (fxy)^2>0 and fxx<0

the given surface f(x,y) has a MAXIMA AT (2,-4)

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