# Why You Should Learn Calculus Photo by Jeswin Thomas on Unsplash

Since its creation, calculus has undoubtedly changed the world of mathematics for the better. It is a brilliant and new way of thinking that explores the principles of continuous change. You may have heard of some of calculus’s many topics, including differentiation, integration, and so on. But have you ever considered how impactful these concepts really are? Many people end up just cramming techniques into their heads, whether it be the power rule or integration by parts, without actually understanding how useful they can be. However, I am here to tell you that calculus has impacts in all sorts of areas, from economics all the way to industrial design. These impacts have heavily aided researchers in their respective areas of work.

# What is Optimization?

In its most basic form, optimization problems involve finding the maximum or minimum of a function. The catch is that in real-world scenarios, there is an added bit of correctly interpreting the obtained optimized values. However, whether you are given a regular mathematics problem inside the classroom or a task of designing the optimal can to be mass-produced, the strategy remains the same.

Step 1: find an appropriate function.

Step 2: optimize the function by finding the critical points or the endpoints.

Step 3: interpret the result in the context of the problem.

Let’s look at an example. Suppose we are given the task of designing a box to hold as much material as possible. The box will be made by cutting out identical squares from all four corners of a 30- by 35-inch sheet of tin. The problem seems difficult, but once we visualize what is actually going on here, it becomes quite simple. Refer to the diagram below.

It is now clear that when the edges fold up to make the 3D box, the box’s dimensions will be x inches by (35–2x) inches by (30–2x) inches. Since 2x can never exceed 30 in the last mentioned dimension, we know that the values of x must be inside of the interval [0 , 15]. Continuing forward, we can represent the function of volume as V(x) = x(35–2x)(30–2x). We can further simplify this function to be V(x) = 4x³-130x²+1050x.

Now, we are looking to find the dimensions of the box that will hold the MOST material. Therefore, we want to find the maximum value of the volume function. To do this, we must first find the first derivative of V(x) and solve for the values of x that make V’(x) = 0. Using the power rule, we can determine that V’(x) = 12x²-260x+1050.

Below are the graphs of V(x), the red line, and V’(x), the blue line.

The value of V’(x) equals 0 at x = 5.369 and x = 16.298. However, note that 16.298 is out of the interval of [0 , 15] that we previously established, so that value can be eliminated instantly; only x = 5.369 is in the interval. In addition, we must also include the endpoints, x = 0 and x = 15, in our analysis as well, since the process of determining extreme values must include both critical points and endpoints (even though it may seem pointless here). A brief summary is provided below:

Critical Point: x = 5.369 inches; V(5.369) = 2509.12 inches³.

Endpoints: x = 0 inches; V(0) = 0 inches³. x = 15 inches; V(15) = 0 inches³.

Clearly, when x = 5.369, the volume function produces the highest value. Now, here comes the most important part: interpreting the solution. What does x = 5.369 actually mean in the context of this task? Well, in our diagram, x represented the length of the squares we cut off from each corner of the tin sheet. Therefore, the final conclusion of this problem should be that if we cut congruent squares with length 5.369 inches from the corners of a 30- by 35-inch sheet of tin, the box formed will hold the maximum volume of material, 2509.12 inches³.

# Optimization in Economics Photo by Sharon McCutcheon on Unsplash

Two areas of economics that calculus affects are the processes of minimizing average cost and maximizing profit. First, we’ll look at how optimization plays a role in minimizing the average cost, and second, we’ll look at its contribution to maximizing profit.

# Minimizing Average Cost

Suppose we have a function C(i) that represents the cost of producing i hundred items. The smallest average cost will be yielded when its derivative is zero. In this situation, the average cost is C(i) / i , which we can call F(i).

Assume we are given C(i) = 6i³-7i²+10i. If such a value exists, how many items need to be produced to yield the smallest average cost?

We have F(i) = C(i) / i, the average cost function, which simplifies to 6i²-7i+10. Differentiating F(i), we find that F’(i) = 12i-7. Setting this function equal to 0, we find that i = 7/12, or roughly 0.583. Note that there is no need to do the first derivative test for a minimum here since we know that the equation for average cost is a quadratic equation that opens up. This means that we’ll need to produce 58.3 items in order to minimize the average cost.

# Maximizing Profit

Similar to the process of minimizing average cost, maximizing profit also involves differentiation. The maximum profit is defined as the production level at which marginal cost equals marginal revenue. Therefore, when dealing with maximum profit, we also need to know the function that models the revenue from selling i items, R(i). This is because the marginal revenue is defined as R’(i), the change in total revenue from an increase in one unit.

As always, let’s look at an example. Imagine that we are given R(i) = 4i and C(i) = 2i³+i²+i, where i represents hundreds of items. Which value of i yields the maximum profit?

R’(i), the marginal revenue, simplifies to 4.

C’(i), the marginal cost, simplifies to 6i²+2i+1.

Setting the two functions equal to each other and solving for i, we determine that i = -0.893 and i = 0.56. Because it does not make sense to have a negative number of items, our final answer is i = 0.56. The production level that maximizes the profit occurs when 0.56 hundred, or 56 items, are produced.

# Final Thoughts

I hope you enjoyed this article regarding some of the contributions calculus has made in the real world. As you can see, calculus is not only useful inside the classroom. Instead, it speaks the fundamental logic of the world we live in, from economic theory to design to statistics to medicine. It may be difficult, but it will be well worth your time to learn.

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