Introduction

Finding extreme values of a function involves identifying the maximum and minimum values the function can achieve. These values can occur at critical points (where the derivative is zero or undefined) or at the boundaries of the domain.

Steps to Find the Minimum Value of a Function

1. Find the derivative of the function:
   Differentiate the given function $f(x)$ with respect to $x$ to obtain $f'(x)$.
2. Solve $f'(x) = 0$:
   Identify the critical points by solving $f'(x) = 0$ or where $f'(x)$ is undefined.
3. Determine the nature of critical points:
   Use the second derivative test or first derivative test:
   • If $f”(x) > 0$ at a critical point, it’s a local minimum.
   • If $f”(x) < 0$, it’s a local maximum.
   • If $f”(x) = 0$, further analysis is needed (e.g., higher derivatives or the first derivative test).
4. Evaluate endpoints (if the domain is closed):
   Compute $f(x)$ at the boundary points of the interval.
5. Compare values:
   Compare the function values at critical points and endpoints to identify the minimum value.

Example: Find the Minimum of $f(x) = x^3 – 6x^2 + 9x + 2$

1. Find the derivative:
   $f'(x) = 3x^2 – 12x + 9$.
2. Solve $f'(x) = 0$:
   $3x^2 – 12x + 9 = 0$
   Divide by 3:
   $x^2 – 4x + 3 = 0$
   Factorize:
   $(x – 1)(x – 3) = 0$
   Critical points: $x = 1$ and $x = 3$.
3. Second derivative test:
   $f”(x) = 6x – 12$.
   At $x = 1$: $f”(1) = 6(1) – 12 = -6$ (local maximum).
   At $x = 3$: $f”(3) = 6(3) – 12 = 6$ (local minimum).
4. Find function values:
   $f(1) = (1)^3 – 6(1)^2 + 9(1) + 2 = 6$.
   $f(3) = (3)^3 – 6(3)^2 + 9(3) + 2 = 2$.
5. Conclusion:
   The minimum value is $2$ at $x = 3$.

Real-Life Application: Minimizing the Cost of Packaging Material

Problem Statement:
A manufacturer wants to design a cylindrical can that holds a fixed volume of $V = 1000 \, \text{cm}^3$. The goal is to minimize the surface area of the can (and hence minimize the cost of the material).

Steps to Solve:

1. Define Variables:
   • Let $r$ = radius of the cylinder’s base (in cm).
   • Let $h$ = height of the cylinder (in cm).
2. Surface Area Formula:
   The surface area $A$ includes the area of two circular bases and the lateral surface area:

   $$A = 2\pi r^2 + 2\pi r h$$

3. Volume Constraint:
   The volume of the cylinder is fixed:

   $$V=\pi r^{2}h$$Substituting $V = 1000$, we get:

   $$h = \frac{1000}{\pi r^2}$$

4. Substitute $h$ in $A$

Replace $h$ in the surface area formula to express $A$ in terms of $r$:

$$A=2\pi r^2+2\pi r\cdot \frac{1000}{\pi r^{\mathrm{2<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; font-size:\; 16px;"></span>}}}$$

Simplify:

$$A = 2\pi r^2 + \frac{2000}{r}$$

1. Minimize $A$ : Find the derivative of $A$ with respect to $r$:

   $$\frac{dA}{dr} = 4\pi r – \frac{2000}{r^2}$$Set $\frac{dA}{dr} = 0$:

   $$4\pi r = \frac{2000}{r^2}$$Multiply through by $r^2$:

   $$4\pi r^3 = 2000$$
   Solve for $r$:

   $$r^3 = \frac{2000}{4\pi} = \frac{500}{\pi}$$ $$r = \sqrt[3]{\frac{500}{\pi}}$$

2. Calculate $h$:
   Substitute $r$ back into the volume constraint:

   $$h = \frac{1000}{\pi r^2}$$

3. Verify with the Second Derivative Test:
   Compute $\frac{d^{2}A}{d{r}^2}$

   $$\frac{d^2A}{dr^2} = 4\pi + \frac{4000}{r^3}$$Since $\frac{d^{2}A}{d{r}^2}>\mathrm{0<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"> </span>}$$A$ has a minimum.

Final Result:

• Using numerical methods (or a calculator), approximate the radius $r$ and height $h$:
  $r \approx 5.42 \, \text{cm}$, $h \approx 10.84 \, \text{cm}$.
• Minimum surface area $A \approx 553.58 \, \text{cm}^2$.

Interpretation:

The optimal dimensions of the can with the least material cost are:

• Radius: $5.42 \, \text{cm}$
• Height: $10.84 \, \text{cm}$.

This ensures the volume requirement is met while minimizing the cost of packaging material.

Real-Life Application in Electrical Engineering: Minimizing Power Loss in a Transmission Line

Problem Statement:
In electrical power transmission, power loss $P_{\text{loss}}$ due to resistance in the transmission lines can be minimized by optimizing the transmission voltage. The goal is to find the optimal transmission voltage $V$ that minimizes the power loss while meeting the system’s constraints.