Question

find the length of the distance from the point (0,0,25/9)to the surface z=xy

Answer 1

The length of the distance from the point (0,0,25/9)to the surface z=xy is $\frac{41}{3}$.

• The length of the perpendicular to the plane passing through the given point is the distance between the point and the plane. In other words, the length of the normal vector dropped from the given point onto the provided plane is the distance between the two objects.
• The shortest perpendicular distance from the point to the specified plane is the distance between the point and the plane. The length of the perpendicular parallel to the normal vector dropped from the provided point to the supplied plane, to put it simply, is the shortest distance between a point and a plane.
• The normal vector's length, which originates at the supplied point and touches the plane, is the shortest distance between a point and a plane.

Now, we need to find the length of the distance from the point (0,0,25/9)to the surface z = xy or, xy - z = 0.

The system of equations resulting from the use of Lagrange multipliers is  resolved as,

2x = λy

2y = λx

2(z − 25/9) = −λ

xy − z = 0

There are a number of options. If x is 0, either y is zero or λ is zero. However, if λ  = 0, then y = 0 and z = 0 according to the second and third equations, respectively. The same is true if any one of x, y, z, or λ is 0—all of them are zero. So the answer is (0, 0, 0).

Divided equations result in x = y and λ = 2 or x = y and λ = 2, depending on whether x, y, z, and λ =0. It follows that ($\frac{4}{3}, \frac{4}{3}, \frac{16}{9}$) and ($\frac{4}{3}, \frac{4}{3}, \frac{16}{9}$) are solutions in the first case, and that the solutions to the second case are incompatible with the fourth equation. A minimum value of $\frac{41}{9}$ is obtained by calculating the f-values of each of the three solutions (corresponding to the second and third solutions). The shortest distance is therefore $\frac{41}{3}$.

Hence, the length of the distance from the point (0,0,25/9)to the surface z=xy is $\frac{41}{3}$.

Learn more here

https://brainly.in/question/9158839

#SPJ9

Answer 2

Answer:

41/3

Step-by-step explanation:

The distance between the point and the plane is equal to the length of the perpendicular to the plane that passes through the specified point. In other words, the distance between the two objects is equal to the length of the normal vector dropped from the given point onto the given plane.

The distance between the point and the plane is the shortest perpendicular distance from the point to the chosen plane. Simply put, the shortest distance between a point and a plane is the length of the perpendicular parallel to the normal vector dropped from the provided point to the provided plane.

The shortest distance between a point and a plane is the length of a normal vector, which starts at the given point and touches the plane.

The system of equations resulting from the use of Lagrange multipliers is  resolved as,

2x = λy

2y = λx

2(z − 25/9) = −λ

xy − z = 0

There are a few choices. Either y is zero or is zero if x is 0. However, the second and third equations, respectively, state that y = 0 and z = 0 if = 0. If any one of x, y, z, or is 0, then everything else is also zero. So, the response is (0, 0, 0).

Depending on whether x, y, z, and = 0, divided equations produce either x = y and = 2 or x = y and = 2. The solutions to the second case are incompatible with the fourth equation, hence it follows that () and () are solutions in the first instance. Calculating the f-values of each of the three solutions yields a minimum value of (corresponding to the second and third solutions).

Hence, the length of the distance from the point (0,0,25/9)to the surface z=xy is .

Learn more here, brainly.in/question/9158839

#SPJ1