Question

Consider the function z=g(x,y)=X³ +13 3 (a) Compute the partial

derivatives of g. gx(x,y) = x² 2-y gy(x,y)=2 - X (b) Find all
critical points of g(x,y). List them in order from smallest x to
largest
8. [4/7 Points] DETAILS PREVIOUS ANSWERS Consider the function z=g(x,y)=X+X-XY+7. (a) Compute the partial derivatives of g. gx(x,y)=x²-y gy(x,y)=2_ -x (b) Find all critical points of g(x,y). List the

Answer 1

The partial derivatives of the function z = g(x,y) are gx(x,y) = 2x - y and gy(x,y) = -x. The function has a single critical point at (0, 0), found by setting the partial derivatives equal to zero and solving for x and y.

Step-by-step explanation:

To address the student's question regarding partial derivatives and critical points, we must evaluate the partial derivatives of the function z = g(x,y) = x + x - xy + 7. There seems to be some confusion in the presentation of the function. Assuming the function z = g(x,y) = x^2 - xy + 7, the partial derivatives are calculated as follows:

• For gx(x,y), which is the partial derivative with respect to x, we get gx(x,y) = 2x - y.
• For gy(x,y), which is the partial derivative with respect to y, we get gy(x,y) = -x.

To find the critical points, we set these partial derivatives equal to zero:

• 2x - y = 0
• -x = 0

Solving these equations, we find that the only critical point is located at (x, y) = (0, 0).

There are no other critical points, and the enumeration by the x-values does not apply as there's only one critical point.

Answer 2

The partial derivatives of the function
`\(g(x, y) = x^3 + (13)/(3)\)` are
`\(g_x(x, y) = 3x^2\) and \(g_y(x, y) = 0\).` The critical point, where both partial derivatives are zero, is (0, 2).

Step-by-step explanation:

Certainly, let's go through the detailed calculations step by step.

Given Information:

`\[ g(x, y) = x^3 + (13)/(3) \]`

Partial Derivatives:

(a) To find the partial derivatives, differentiate
`\(g\)` with respect to
`\(x\) and \(y\):`

`\[ g_x(x, y) = (\partial g)/(\partial x) = 3x^2 \]`

`\[ g_y(x, y) = (\partial g)/(\partial y) = 0 \]`

(b) To find the critical points, set
`\(g_x\) and \(g_y\)` equal to zero and solve for
`\(x\) and \(y\):`

`\[ 3x^2 = 0 \implies x = 0 \]`

`\[ 2 - x = 0 \implies x = 2 \]`

Critical Point:

The critical point is where both partial derivatives are zero. Combining the solutions, we get
`\( (x, y) = (0, 2) \).`

(a) The partial derivatives
`\(g_x\) and \(g_y\)` are obtained by differentiating
`\(g(x, y)\)` with respect to
`\(x\) and \(y\),` respectively. The derivative of
`\(x^3\)` with respect to
`\(x\) is \(3x^2\)` , and the constant term
`\((13)/(3)\)` differentiates to zero. Therefore,
`\(g_x(x, y) = 3x^2\) and \(g_y(x, y) = 0\).`

(b) To find the critical points, we set
`\(g_x\) and \(g_y\)` equal to zero. For
`\(g_x\), \(3x^2 = 0\)` implies
`\(x = 0\). For \(g_y\), \(2 - x = 0\)` results in
`\(x = 2\).` Combining these solutions, the critical point is
`\( (0, 2) \),` where both partial derivatives are zero.

In summary, the partial derivatives
`\(g_x(x, y) = 3x^2\) and \(g_y(x, y) = 0\).` The critical point is
`\( (0, 2) \),` and these calculations provide a detailed understanding of the function's behavior and critical characteristics.