Find the point(s) on the surface z=xy+1x+1y which the tangent plane is horizontal.

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Alyvia Marks · Accepted Answer

Considering: The surface z=xy+1x+1yTo find the point (s) on the surface z=xy+1x+1y at which the tangent plane is horizontal.Use:If the tangent plane is horizontal, the gradient must be zero in the z - direction.Therefore the x and y component are zero.Let z(x,y)=xy+1x+1yTo find ∂z∂xCalculate the above equation as a function of x.To get,∂z∂x=∂∂x(xy+1x+1y)=y−1x2Then, ∂z∂x=y−1x2To findDifferentiate the above equation with respect to y,To get,PSKfrac{partial z}{partial y} = frac{partial}{partial x} (xy + frac{1}{x} + frac{1}{y})ZSK=y−1y2}Then, ∂z∂y=x−1y2By u sin g if the tangent plane is horizontal, the gradient must be zero in the z - direction.So set the pertial derivative with respect to x and y equal to zero.To get,∂z∂x=y−1x2=0∂z∂y=x−1y2=0Consider x−1y2=0Add on both side by frac{1}{y^2}ZSKTo get,x=1y2Plug x=1y2∈y−1x2=0To get,y−1(1y2)2=0y−11y4=0y−y41=0To get,y−y4=0y(1−y3)=0That is,y=0,1−y3=0Solve 1−y3=0fory.Add on both side by y3,To get,1=y3Taking cube root on both side,To get,y = 1Plug y=1∈x=1y2To get,x=112=1To get,x = 1 and y =1Plug the value of x = 1 and y = 1 in the given equation to find the point 3rd coordinate.To get,z(1,1)=(1)(1)+11+11= 1 + 1 + 1= 3To get,z(1,1) = 3Then,(x, y, z) = (1, 1, 3)Therefore,The point (1,1,3)onthesurfacez=xy+1x+1y at which the tangent plane is horizontal.

Find the point(s) on the surface z = xy+1/x +1/y which the tangent plane is horizontal.

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