Optimization problems deal with finding a mathematical function's maximum or minimum. In this post, let's explore how to tackle these real-world problems.
An application of differential calculus involves finding the answers to some of these questions:
What is the highest distance an object can reach in a projectile?
What is the maximum box space I can make with paper?
How many persons should there be in a production line to gain the highest revenue?
Most of the questions deal with finding the maximum (or minimum) of something. We call these situations optimization problems. It is one of the elementary applications of differential calculus.
In dealing with these types of problems, we can summarize it in these steps:
Identify the Problem
Model Using a Function
Solve for Extrema
Evaluate Results
Example
For illustration purposes, we will answer a sample optimization problem: what is the largest open square box we can fold from a 700 sq. cm piece of paper?
How do we start or find a way to answer that mathematically?
Identify the Problem
The first step is to identify what is required. In this problem, given a sheet of paper, we need to find the giant open box we can make with it. We can translate that problem as finding the maximum volume of a box given a paper of fixed area.
Let's draw the open box and define some variables:
\(x\) as the length of the square base
\(y\) as the height of the box.
Now, let's fold the paper to its shape. Since we have a paper of 700 sq. cm, the total surface area of the open box, when folded, must also be 700 sq. cm.
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After identifying what is required, we create a model for it. We need to relate the volume of the open box with one of the dimensions of it. We start with identifying the primary expression for finding volumes of boxes. According to Cavalieri's theorem: \(V=Ah\).
\(V\) is the volume of the open box
\(A\) is the cross-sectional area of the object
\(h\) is the height of the object
We can develop this equation using our predefined variables \(x\) and \(y\). Expressing it in terms of these, we have \(V=x^2y\).
There are two variables, \(x\) and \(y\). Although we can use this expression, we must find a way to express the primary function in terms of a single variable only (either \(x\) or \(y\)).
To do this, we make a secondary expression which we can get from its surface area, \(s\). For an open box, the variable \(s\) is the sum of the base area and four sides: \(s = x^2+4xy\). From our problem, this must equal 700 sq. cm.
\(s = x^2+4xy\)
\(700 = x^2+4xy\)
We can express one of the variables with the other. In our case, we isolate variable y and insert it into the primary function.
\(700 = x^2+4xy\)
\(y=\frac{175}{x}-\frac{x}{4}\)
\(V=x^2\left(\frac{175}{x}-\frac{x}{4}\right)\)
This resulting equal is what we call the main function. We will use this expression to find the answer to our problem.
Solve for Extrema
Model
Now that we have the function between the box's volume \(V\) and the length of the square base \(x\), we can solve the problem. To start, let's plot this primary function.
From the graph, there corresponds a \(x\) value where \(V\) is at its highest - or the maxima of the function. Our task is to find this coordinate.
Let's recall that an extrema has a slope of zero. With this fact, we get the first derivative (slope) of the function:
\(V=x^2\left(\frac{175}{x}-\frac{x}{4}\right)\)
\(V^{\prime}(x)=175-\frac{3 x^2}{4}\)
Afterward, we equate the first derivative to zero to find the values for the critical number, \(c\).
\(V^{\prime}(x)=175-\frac{3 x^2}{4}=0\)
\(x=15.275,-15.275\)
There are two \(x\) values which means that two possible length values of the square box would give the maximum volume \(V\), which are 15.275 and -15.275. Between the two results, there is no negative length; hence, the square box must have a distance of 15.275 units.
Evaluate Results
Now that we have \(x\) we can solve for the height of the box \(y\) and the volume of the box \(V\).
We can answer the question: "What is the largest open square box we can make from a 700 sq. cm piece of paper?" In this case, it would be a box having a volume of 1782.11 cubic centimeters (0.78L). We can achieve a box of this size by having a square base of 15.275 cm and a height of 7.638 cm.
To verify our answer, the total surface area \(s\) must equal 700 sq. cm, as stated in the problem. Evaluating our expression for \(s\), it is:
\(s=x^2+4 x y=15.275^2+4(15.275)(7.638)\)
\(s=700 \mathrm{~cm}^2\)
Summary
Optimization problems deal with finding a function's maximum or minimum.
In dealing with these types of problems, we can summarize them in these steps: (1) identify the problem, (2) model using a function, (3) solve for extrema, and (4) evaluate results.
Identifying the problem involves finding what is required and assigning variables.
Modeling involves finding the primary connection among the variables. We use math functions to accomplish this.
Solving for the extrema involves finding the maximum or minimum of the modeled function. We equate the function's first derivative with zero and solve for the unknowns.
After solving, we use the obtained results and evaluate if these values are the solution to the problem.
