Given the equation:

$z = y^2 + x^3 + 5x + 9$ and the relationship $y = 5 - x$, we can solve the problem in two parts.

Part a) Find the possible values of $x$ for which $z$ has a stationary value

Step 1: Substitute $y = 5 - x$ into the equation for $z$

$z = (5 - x)^2 + x^3 + 5x + 9$

Expanding $(5 - x)^2$:

$z = (25 - 10x + x^2) + x^3 + 5x + 9$

Simplifying:

$z = x^3 + x^2 - 10x + 5x + 25 + 9$ $z = x^3 + x^2 - 5x + 34$

Step 2: Find the derivative of $z$ with respect to $x$ (since $z$ depends only on $x$ now)

$\frac{dz}{dx} = 3x^2 + 2x - 5$

Step 3: Set the derivative equal to zero to find the critical points (stationary values)

$3x^2 + 2x - 5 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 3$, $b = 2$, and $c = -5$.

$x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-5)}}{2(3)}$ $x = \frac{-2 \pm \sqrt{4 + 60}}{6}$ $x = \frac{-2 \pm \sqrt{64}}{6}$ $x = \frac{-2 \pm 8}{6}$

The two solutions are:

$x = \frac{-2 + 8}{6} = \frac{6}{6} = 1$ $x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3}$

So, the possible values of $x$ for which $z$ has stationary values are $x = 1$ and $x = -\frac{5}{3}$.

Part b) Find the minimum value of $z$

Step 1: Evaluate $z$ at the stationary points

We already have the expression for $z$:

$z = x^3 + x^2 - 5x + 34$

Now, calculate $z$ at $x = 1$ and $x = -\frac{5}{3}$.

• For $x = 1$:

$z = 1^3 + 1^2 - 5(1) + 34 = 1 + 1 - 5 + 34 = 31$

• For $x = -\frac{5}{3}$:

$z = \left(-\frac{5}{3}\right)^3 + \left(-\frac{5}{3}\right)^2 - 5\left(-\frac{5}{3}\right) + 34$ $z = -\frac{125}{27} + \frac{25}{9} + \frac{25}{3} + 34$

Simplifying the terms:

$z = -\frac{125}{27} + \frac{75}{27} + \frac{225}{27} + \frac{918}{27}$ $z = \frac{-125 + 75 + 225 + 918}{27} = \frac{1093}{27} \approx 40.48$

Step 2: Compare the values

• $z(1) = 31$
• $z\left(-\frac{5}{3}\right) \approx 40.48$

Thus, the minimum value of $z$ occurs at $x = 1$, and the minimum value of $z$ is $31$.

Final Answer:

a) The possible values of $x$ for which $z$ has a stationary value are $x = 1$ and $x = -\frac{5}{3}$.

b) The minimum value of $z$ is $31$.

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Do you want further details on the calculations? Here are five related questions for further exploration:

1. How would the stationary points change if the equation of $z$ had a different cubic term?
2. What is the geometric interpretation of the stationary points found?
3. Can you confirm whether the critical points are minima or maxima using the second derivative test?
4. How does the stationary value change if the constant term in $z$ is modified?
5. What is the significance of stationary points in optimization problems?

Tip: The second derivative test can be a useful tool to determine if a critical point is a minimum, maximum, or inflection point.