Question

Given f(x, y) = 15x 3 − 3xy 15y 3 , find all points at which fx(x, y) = fy(x, y) = 0 simultaneously

Answer 1

The function
`\(f(x, y) = 15x^3 - 3xy + 15y^3\)` has no points where both
`\(f_x(x, y)\)` and
`\(f_y(x, y)\)` are equal to zero simultaneously.

Explanation:

To find the critical points where both partial derivatives are zero, we compute
`\(f_x\) and \(f_y\).`

`\(f_x = 45x^2 - 3y\)`

`\(f_y = -3x + 45y^2\)`

Setting both equations to zero to find critical points:

`\(45x^2 - 3y = 0\)`

`\(-3x + 45y^2 = 0\)`

From the first equation,
`\(y = 15x^2\)`. Substituting this into the second equation gives
`\(-3x + 45(15x^2)^2 = 0\)`

Simplifying, we get
`\(3x(1 - 2250x^4) = 0\)`

This implies
`\(x = 0\) or \(x^4 = (1)/(2250)\)`

The value
`\(x = 0\)` yields
`\(y = 0\)` from
`\(y = 15x^2\)`, giving a point (0, 0).

For
`\(x^4 = (1)/(2250)\)`, there are no real solutions within the domain.

Hence, the function
`\(f(x, y)\)` has only one critical point at (0, 0), where neither
`\(f_x\) nor \(f_y\)` are zero. This implies that there are no points where both partial derivatives are zero simultaneously.

Answer 2

The function f(x, y) = 15x^3 - 3xy + 15y^3 requires finding where both partial derivatives are zero; solving the system of partial derivatives gives multiple solutions including (0, 0).

Step-by-step explanation:

The student has a function f(x, y) = 15x^3 − 3xy + 15y^3 and needs to find all points where the partial derivatives with respect to x and y are both zero. Calculating these partial derivatives, we get:

• f_x(x, y) = ∂f/∂x = 45x^2 - 3y
• f_y(x, y) = ∂f/∂y = -3x + 45y^2

To find the points where both derivatives are zero, we set them equal to zero and solve the system of equations:

• 45x^2 - 3y = 0
• -3x + 45y^2 = 0

Solving for x and y, we find that this system has multiple solutions including the point (0, 0), among others.