Optimization Problems

You are tasked with enclosing a rectangular field with a fence. You are given 400 ft of fencing materials. However, there is a barn on one side of the field (thus, fencing is not required on one side of the rectangular field). What dimensions of the field will produce the largest area subject to the 400 ft of fencing materials?

We will solve this problem using the method outlined in the article.

Step 1: Fully understand the problem

Let's draw the important information out from the problem.

We need to fence three sides of a rectangular field such that the area of the field is maximized. However, we only have 400 ft of fencing material to use. Thus, the perimeter of the rectangle must be less than or equal to 400 ft.

Step 2: Draw a diagram

Clearly, you don't have to be an artist to sketch a diagram of the problem!

The diagram of the fencing problem helps us to better visualize the problem - StudySmarter Original

Step 3: Introduce necessary variables

Looking at the diagram above, we've already introduced some variables. We'll let the height of the rectangle be represented by $h$. We'll let the width of the rectangle be represented by $w$.

$$\begin{gathered} h=height \\ w=width \end{gathered}$$

So, we can calculate area and perimeter as

$$\begin{gathered} Area=h\times w \\ Perimeter=h+2w \end{gathered}$$

Step 4: Set up the problem by finding relationships within the problem

The fencing problem wants us to maximize area $A$, subject to the constraint that the perimeter $P$ must be greater or less than 400 ft. Intuitively, we know that we should use all 400 ft of fencing to maximize the area.

So, our problem becomes:

$$\begin{gathered} maximizeA=h\times w \\ suchthatP=400=h+2w \end{gathered}$$

Since we seek to maximize the area, we must write the area in terms of the perimeter to achieve one single equation. In this example, we will write the area equation in terms of width, $A\left(w\right)$.

First, let's solve for the height, $h$:

$$\begin{gathered} 400=h+2w \\ h=400-2w \end{gathered}$$

Now, plug into the area in terms of the width equation, $A\left(w\right)$

$\begin{array}{rcl}A\left(w\right)& =& (400-2w)\left(w\right)\\ & =& 400w-2w^2\end{array}$

Step 5: Find the absolute extrema

Now that we have a single equation containing all of the information from the problem, we want to find the absolute maximum of $A\left(w\right)$. We can define an interval for w so we can use the Closed Interval Method.

For starters, we know that w cannot be smaller than 0. If we let $h=0$, according to our perimeter equation, we have

$\begin{array}{rcl}P& =& h+2w\\ 400& =& 2w\\ w& =& 200\end{array}$

This tells us that if $h=0$, the maximum width possible is 200. So our closed interval for $w$ is $[0,200]$.

To apply the Closed Interval Method:

First, find the extrema of $A\left(w\right)$ by taking the derivative and setting it equal to 0.

$\begin{array}{rcl}A\text{'}\left(w\right)& =& 400-4w\\ 0& =& 400-4w\\ 4w& =& 400\\ w& =& 100\end{array}$

Second, plug in the critical values $w=0,w=100,andw=200intoA\left(w\right)$ and identify the largest area.

$$\begin{gathered} A\left(0\right)=400\left(0\right)-2\left(0^2\right) \\ =0 \end{gathered}$$

$$\begin{gathered} A\left(100\right)=400\left(100\right)-2\left({100}^2\right) \\ =20000 \end{gathered}$$

$$\begin{gathered} A\left(200\right)=400\left(200\right)-2\left({200}^2\right) \\ =0 \end{gathered}$$

So, the largest value of $A$ occurs at $w=100$ where $A=20,000f{t}^2$.

We can confirm this using the First Derivative Test.

Graphing $A\text{'}\left(w\right)$...

We can apply the First Derivative Test to the graph of the derivative - StudySmarter Original

$A\text{'}\left(w\right)$ clearly only equals 0 at one point, $w=100$. For all $c<100$, $A\text{'}\left(w\right)$ is positive (above the x-axis). For all $c>100$, $A\text{'}\left(w\right)$ is negative (below the x-axis). So, by the First Derivative Test, $w=100$ is the absolute maximum of $A\left(w\right)$.

Let's plug in $w=100$ to our perimeter equation to find out what h should be.

$$\begin{gathered} 400=h+2\left(100\right) \\ h=200 \end{gathered}$$

Therefore, to maximize the area enclosed by the fence subject to our material constraints, we should use a rectangle with a width of 100 ft and a height of 200 ft.

You are tasked with building a can that holds 1 liter of liquid. To maximize profit, you must build the can such that the material used to build it is minimized. What is the minimum surface area of the can required?

Again, we will solve this problem using the method outlined in the article.

Step 1: Fully understand the problem

Let's draw the important information out from the problem.

We need to build a can that holds 1 liter of liquid while minimizing the material used to build it. Essentially, this means we need to minimize the can's surface area.

Step 2: Draw a diagram

With this diagram, we can better understand what the problem is asking us to do.

The diagram of the can problem helps us to better visualize the problem - StudySmarter

Step 3: Introduce necessary variables

Looking at the diagram above, we've already introduced some variables. We'll let the radius of the cylindrical can be represented by $r$. We'll let the height of the cylinder be represented by $h$. So, the volume of the cylinder $V$ is $V={\mathrm{\pi r}}^2\mathrm{h}$ and the surface area of the cylinder $A$ is $A=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^2$.

Step 4: Set up the problem by finding relationships within the problem

The can problem wants us to minimize the surface area $A$ subject to the constraint that the can must hold at least 1 liter. Intuitively, we know that to minimize surface area, we should build a can that holds 1 liter of liquid. However, since we are looking for a length measurement for $r$ and $h$, we should convert liters into cubic centimeters. Thus, we should build a can that holds 1,000 cm³ of liquid.

So, our problem becomes:

$$\begin{gathered} minimizeA=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^2 \\ \mathrm{subject}\mathrm{to}V=1000={\mathrm{\pi r}}^2\mathrm{h} \end{gathered}$$

Since we seek to minimize the surface area, we must write the area in terms of the volume to achieve one single equation.

First, let's solve for $h$:

$$\begin{gathered} 1000={\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{h}=\frac{1000}{{\mathrm{\pi r}}^2} \end{gathered}$$

Now, plug into the area equation:

$$\begin{gathered} A=2\mathrm{\pi r}\left(\frac{1000}{{\mathrm{\pi r}}^2}\right)+2{\mathrm{\pi r}}^2 \\ \mathrm{A}=\frac{2000}{\mathrm{r}}+2{\mathrm{\pi r}}^2 \end{gathered}$$

Step 5: Find the absolute extrema

Now that we have a single equation containing all the information from the problem, we want to find the absolute minimum of $A$.

We know that $r>0$. However, we do not have an upper bound for $r$.

First, we'll find the extrema of $A$ by taking the derivative and setting it equal to 0.

$$\begin{gathered} A\text{'}=4\mathrm{\pi r}-\frac{2000}{r^2} \\ 0=4\mathrm{\pi r}-\frac{2000}{r^2} \end{gathered}$$

Graphing the derivative:

We can apply the first derivative test to the graph of the derivative - StudySmarter Original

We can see $A\text{'}=0$ at one point. We can confirm that the point $r=5.4192608391249$ is an absolute minimum for $A$ by applying the First Derivative Test. Looking at the graph, For all $c<5.4192608391249$, $A\text{'}\left(r\right)$ is negative (below the x-axis). For all $c>5.4192608391249$, $A\text{'}\left(w\right)$ is positive (above the x-axis). So, by the First Derivative Test, $r=5.4192608391249$ is the absolute maximum of $A\left(r\right)$.

Let's plug in $r=5.4192608391249$ to our volume equation to find out with $h$ should be.

$$\begin{gathered} 1000=\pi {(5.4192608391249)}^2\mathrm{h} \\ \mathrm{h}=10.8385208518578 \end{gathered}$$

So, to build a can that holds at least 1 liter, the minimum surface area required is

$$\begin{gathered} A=2\pi (5.4192608391249)(10.8385208518578)+2\pi {(5.4192608391249)}^2 \\ A=553.58c{m}^2 \end{gathered}$$