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

User Alanl
by
8.1k points

2 Answers

4 votes

Final Answer:

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.

User JoshAdel
by
8.9k points
4 votes

Final answer:

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.

User Apirogov
by
8.7k points
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