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We have the following function

z=-x³/12-4 y²+6 x y+20 x-2 y
(a) Find the stationary values. That is, solve the first order conditions for the stationary values

1 Answer

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**Identification of Elements:**

1. **Start/End:** Oval shape representing the start and end of the flowchart.

2. **Process:** Rectangle indicating a process or computation step.

3. **Input/Output:** Parallelogram representing input or output operations.

4. **Decision:** Diamond shape for decision-making points.

5. **Arrow:** Arrows connecting the elements to show the flow of the process.

**Flowchart:**

1. **Start:** Oval

2. **Input:** Rectangle (Input x, y)

3. **Process:** Rectangle (Calculate z=-x³/12-4y²+6xy+20x-2y)

4. **Output:** Parallelogram (Display z)

5. **Decision:** Diamond (Check if first-order conditions are met)

6. **Process:** Rectangle (If yes, display stationary values; if no, end)

7. **End:** Oval

**Pseudo Code:**

```

Start

Input x, y

z = -x³/12 - 4y² + 6xy + 20x - 2y

Display z

If first-order conditions are met:

Display stationary values

Else:

End

End

```

Final Answer

Stationary values are found by solving the first-order conditions.

Step-by-step explanation

To find stationary values, we need to calculate the first-order partial derivatives of the given function with respect to x and y. Setting these derivatives equal to zero and solving for x and y will provide the stationary points. Let
\( (\partial z)/(\partial x) = -(x^2)/(4) + 6y + 20 \) and \( (\partial z)/(\partial y) = -8y + 6x - 2 \).Setting both of these equal to zero and solving simultaneously will yield the values of x and y at the stationary points.

For example, solving
\( -(x^2)/(4) + 6y + 20 = 0 \) and \( -8y + 6x - 2 = 0 \) simultaneously will give the values of x and y. Once these values are found, they can be substituted back into the original function to obtain the corresponding z values at the stationary points.

In summary, the stationary values are obtained by solving the system of equations formed by setting the partial derivatives of the function equal to zero.

User David Kuhta
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