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

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Best Answer

Given that:The surface z = xy+(1)/(x) +(1)/(y)

To find the point (s) on the surface z = xy+(1)/(x) +(1)/(y) 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 + (1)/(x) + (1)/(y)

To find (∂z)/(∂ x)

Differentiate the above equation with respect to x,

To get,

(∂z)/(∂x) = (∂)/(∂ x) (xy + (1)/(x) + (1)/(y))

= y - (1)/(x^2)

Then,

(∂ z)/(∂l x) = y - (1)/(x^2)

To find

Differentiate the above equation with respect to y,

To get,

(∂ z)/(∂ y) = (∂)/(∂ x) (xy + (1)/(x) + (1)/(y))

= y - (1)y^2)

Then,

(∂ z)/(∂ y) = x - (1)/(y^2)

By 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 - (1)/(x^2) = 0

(∂z)/(∂y) = x - (1)/(y^2) = 0

Consider x - (1)/(y^2) = 0

Add on both side by (1)/(y^2)

To get,

x = (1)/(y^2)

Plug x = (1)/(y^2) in y - (1)/(x^2) = 0

To get,

y - (1)/(((1)/(y^2))^2) = 0

y - (1)/((1)/(y^4)) = 0

y - (y^4)/(1) = 0

To get,

y - y^4 = 0

y(1 - y^3) = 0

That is,

y = 0, 1 - y^3 = 0

Solve 1 - y^3 = 0 for y.

Add on both side by y^3,

To get,

1 = y^3

Taking cube root on both side,

To get,

y = 1

Plug y = 1 in x = (1)/(y^2)

To get,

x = (1)/(1^2) = 1

To get,

x = 1 and y =1

Plug 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) + (1)/(1) + (1)/(1)

= 1 + 1 + 1

= 3

To get,

z(1,1) = 3

Then,

(x, y, z) = (1, 1, 3)

Therefore,

The point (1,1,3) on the surface z = xy + (1)/(x) + (1)/(y) at which the tangent plane is horizontal.

To find the point (s) on the surface z = xy+(1)/(x) +(1)/(y) 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 + (1)/(x) + (1)/(y)

To find (∂z)/(∂ x)

Differentiate the above equation with respect to x,

To get,

(∂z)/(∂x) = (∂)/(∂ x) (xy + (1)/(x) + (1)/(y))

= y - (1)/(x^2)

Then,

(∂ z)/(∂l x) = y - (1)/(x^2)

To find

Differentiate the above equation with respect to y,

To get,

(∂ z)/(∂ y) = (∂)/(∂ x) (xy + (1)/(x) + (1)/(y))

= y - (1)y^2)

Then,

(∂ z)/(∂ y) = x - (1)/(y^2)

By 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 - (1)/(x^2) = 0

(∂z)/(∂y) = x - (1)/(y^2) = 0

Consider x - (1)/(y^2) = 0

Add on both side by (1)/(y^2)

To get,

x = (1)/(y^2)

Plug x = (1)/(y^2) in y - (1)/(x^2) = 0

To get,

y - (1)/(((1)/(y^2))^2) = 0

y - (1)/((1)/(y^4)) = 0

y - (y^4)/(1) = 0

To get,

y - y^4 = 0

y(1 - y^3) = 0

That is,

y = 0, 1 - y^3 = 0

Solve 1 - y^3 = 0 for y.

Add on both side by y^3,

To get,

1 = y^3

Taking cube root on both side,

To get,

y = 1

Plug y = 1 in x = (1)/(y^2)

To get,

x = (1)/(1^2) = 1

To get,

x = 1 and y =1

Plug 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) + (1)/(1) + (1)/(1)

= 1 + 1 + 1

= 3

To get,

z(1,1) = 3

Then,

(x, y, z) = (1, 1, 3)

Therefore,

The point (1,1,3) on the surface z = xy + (1)/(x) + (1)/(y) at which the tangent plane is horizontal.