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.