Optimization Problems in Calculus

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Optimization Problems
• 2. Formula Used in Optimization Problems
• 3. Solved Examples
• 4. Conclusion

Optimization Problems:
Optimization problems involve finding the maximum or minimum values of a function subject to certain conditions. In calculus, optimization is used to find values such as the largest or smallest value of a function in a given domain, often related to real-world problems in physics, economics, engineering, and more.

To solve optimization problems, we typically follow these steps:

1. Find the first derivative of the function (representing the rate of change).
2. Set the derivative equal to zero to find critical points (potential maxima or minima).
3. Test the critical points using the second derivative or other methods to determine if they correspond to a maximum, minimum, or saddle point.

Example 1: Maximizing Profit

Suppose the cost function for manufacturing a product is given by $C(x) = 4x^2 - 8x + 15$, and the revenue function is $R(x) = 10x - x^2$, where $x$ is the number of units produced and sold. Find the number of units that maximizes profit.

Solution:

The profit function is given by:

$P(x) = R(x) - C(x) = (10x - x^2) - (4x^2 - 8x + 15)$

Simplifying:

$P(x) = 10x - x^2 - 4x^2 + 8x - 15$

$P(x) = -5x^2 + 18x - 15$

Now, differentiate the profit function:

$P'(x) = \dfrac{d}{dx} (-5x^2 + 18x - 15)$

$P'(x) = -10x + 18$

To find critical points, set $P'(x) = 0$:

$-10x + 18 = 0$

$x = \frac{18}{10} = 1.8$

Now, use the second derivative to confirm that this is a maximum:

$P''(x) = -10$

Since $P''(x) < 0$, we have a local maximum at $x = 1.8$.

Thus, the number of units that maximizes profit is: $\boxed{1.8}$.

Example 2: Minimizing the Surface Area of a Box

A box is to be constructed with a square base and an open top. The volume of the box must be $32 \, \text{m}^3$. Find the dimensions of the box that minimize the surface area.

Solution:

Let the side of the square base be $x$ and the height of the box be $h$.

The volume is given by:

$V = x^2h = 32$

Thus, $h = \dfrac{32}{x^2}$.

The surface area is given by:

$A = x^2 + 4xh$

Substitute $h = \dfrac{32}{x^2}$ into the surface area formula:

$A = x^2 + 4x .\left( \dfrac{32}{x^2} \right)$

$A = x^2 + \dfrac{128}{x}$

Now, differentiate the surface area function:

$A'(x) = 2x - \dfrac{128}{x^2}$

Set $A'(x) = 0$:

$2x = \dfrac{128}{x^2}$

$2x^3 = 128$

$x^3 = 64$

$x = 4$

Substitute $x = 4$ into the volume equation to find $h$:

$h = \dfrac{32}{4^2} = \dfrac{32}{16} = 2$

Thus, the dimensions of the box that minimize surface area are:

$\boxed{4 \, \text{m} \times 4 \, \text{m} \times 2 \, \text{m}}$.

Example 3: Maximizing Area in Geometry

Find the dimensions of a rectangle with a perimeter of $30 \, \text{m}$ that maximizes the area.

Solution:

Let the length of the rectangle be $l$ and the width be $w$. The perimeter is given by:

$P = 2l + 2w = 30$

Thus, $l = 15 - w$.

The area is given by:

$A = lw = (15 - w)w = 15w - w^2$

Now, differentiate the area function:

$A'(w) = 15 - 2w$

Set $A'(w) = 0$:

$15 - 2w = 0$

$w = 7.5$

Substitute $w = 7.5$ into the perimeter equation to find $l$:

$l = 15 - 7.5 = 7.5$

Thus, the dimensions of the rectangle that maximize the area are:

$\boxed{7.5 \, \text{m} \times 7.5 \, \text{m}}$.

Example 4: Minimizing Cost in Manufacturing

A company produces a product where the cost is given by $C(x) = 500 + 3x + 0.1x^2$, where $x$ is the number of items produced. Find the number of items that minimizes the cost.

Solution:

The cost function is:

$C(x) = 500 + 3x + 0.1x^2$

Differentiate the cost function:

$C'(x) = 3 + 0.2x$

Set $C'(x) = 0$:

$3 + 0.2x = 0$

$x = -\frac{3}{0.2} = -15$

Since the number of items cannot be negative, this critical point indicates a minimum cost at $x = 15$.

Thus, the number of items that minimizes the cost is: $\boxed{15}$.