The function f(x,y) = 2x2 +2xy – y3 hasa)Only one stationary point at (0,0) b)Two stationary points at (0,0) and (1/6, -1/3) c)Two stationary points at (0,0) and (1,-1) d)No stationary pointCorrect answer is option 'B'. Can you explain this answer?

Question

The function f(x,y) = 2x2 +2xy &#8211; y3 hasa)Only one stationary point at (0,0) b)Two stationary points at (0,0) and (1/6, -1/3) c)Two stationary points at (0,0) and (1,-1) d)No stationary pointCorrect answer is option 'B'. Can you explain this answer?

Answer

Answer

Answer

Explanation:

The given function is f(x,y) = 2x^2 - 2xy + y^3.

To find the stationary points of the function, we need to find the values of x and y for which the partial derivatives of f with respect to x and y are equal to zero.

Taking the partial derivative of f with respect to x, we get:

∂f/∂x = 4x - 2y

Setting this equal to zero and solving for x, we get:

4x - 2y = 0
4x = 2y
x = y/2

Taking the partial derivative of f with respect to y, we get:

∂f/∂y = -2x + 3y^2

Setting this equal to zero and solving for y, we get:

-2x + 3y^2 = 0
-2x = -3y^2
x = (3/2)y^2

Now, we can substitute the value of x in terms of y into the equation x = y/2 to find the values of x and y for the stationary points.

(3/2)y^2 = y/2
3y^2 = y
3y^2 - y = 0
y(3y - 1) = 0

This gives us two possible solutions for y:
1) y = 0
2) 3y - 1 = 0
3y = 1
y = 1/3

Substituting these values back into the equation x = y/2, we get the corresponding values of x:
1) For y = 0, x = 0/2 = 0
2) For y = 1/3, x = (1/3)/2 = 1/6

Therefore, the stationary points of the function are (0,0) and (1/6, 1/3).

Answer:

The correct answer is option B, which states that there are two stationary points at (0,0) and (1/6, 1/3).

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