# Applications of Differentiation – Applied Optimization – Application of Maximums and Minimums

So far, we have used derivatives to find the maximum and minimum values of a function. However, in the real world, you will be asked to maximize or minimize the value of something. Here is an example of an optimization problem:

You have about 100  feet of fence materials and you want to build three sides of a rectangular fence with a wall forming the fourth side. What is the maximum possible area (square footage) of the fence that you can build?

The process of finding maxima or minima is called optimization. The function we’re optimizing is called the objective function. In this case, the area is the objective function. In the same problem, we are given a limit. The limit is the amount of fencing material that will make up the perimeter. Therefore, the constraint equation is the perimeter.

For most optimization problems, there will be two (or more) variables, or unknowns, and two equations (the optimization equation and the constraint equation). Essentially, you have a system of equations that can be solved by using the substitution method.

 Guidelines for Solving Optimization Problems Translate the word problem into a picture. This will help you to visualize any shapes, and write the appropriate dimensions. Determine the objective function. Look for words indicating the largest or smallest value of something. Determine the constraint equation. Solve the equation sign for one of the variables and substitute inside the objective function. Now you have a function of only one variable. Perform the first and second derivative tests on the combined objective function. Take the first derivative. Set the first derivative equal to zero to find the critical points. Take the second derivative of the function and substitute the critical points into it. If the result is positive, it is a local min. If the result is negative, then it is a local max. Plug in critical numbers into the original objective function to see which one gives you the minimum or maximum value (depends on question asked). Look back at the question to make sure you answered what was asked.

Example 1:

You have about 100 feet of fence materials and you want to build three sides of a rectangular fence with a wall forming the fourth side. What is the maximum possible area (square footage) of the fence that you can build?

Solution:

Translate the word problem into a picture. This will help you to visualize any shapes and write the appropriate dimensions.

Sketch according to the given statement: Determine the objective function. Look for words indicating the largest or smallest value of something.

Since we are trying to maximize the area, the area will be the objective function.

Let’s assign the length a variable, x, and the width a variable, y:

Objective function:  A = (y) (x)

Determine the constraint equation.

Our constraint is the amount of fencing material that will make the perimeter of the fencing. The fourth wall will not count, so we only have to add the three sides. The perimeter is therefore:

100 = y + 2x

Let’s isolate y in this equation.

y = 100 – 2x

Let’s put this equation into the objective function.

A (x) = (100−2x) x = 100x−2x2

Perform the first and second derivative test on the combined objective function and test the critical points.

Our combined objective function is A (x) = 100x−2x2.

By the power rule, the derivative is A’(x) = 100−2x

Set the derivation equal to 0 to find the critical points.

0 = 100−4x

4x = 100

x = 25

Our only critical point is x = 25.

Let’s take the second derivative of the function:

If A’(x) = 100−2x,

A’’(x) = -2

Since the second derivative is negative for any x value, then the entire function of A(x) is concave down (looks like a frown). This means that at x = 25, there is a maximum value.

Let’s substitute x = 25 into the constraint equation:

y = 100 – 2x

y = 100 – 2(25)

y = 50

Look back at the question to make sure you answered what was asked.

So, the dimensions of the field which is equal to the largest area, will be 25 ft by 50 ft:

A = 50 ft * 25 ft = 1250 square foot

Example 2:

Let’s consider that there is a person who wants to make a cylinder to hold 1000 ml of the liquid in it. Find the dimensions (in cm^3) used to minimize the construction material used in it.

Solution:

Translate the word problem into a picture. This will help you to visualize any shapes and write the appropriate dimensions.

Let’s sketch the cylinder. ‘r’ is the radius of the cylinder and ‘h’ is the height of the cylinder.

Determine the objective function. Look for words indicating the largest or smallest value of something.

Since we are minimizing the amount of material used, our objective function will be the surface area.

The equation for the surface area of a cylinder is:

S = 2πrh + 2πr2 (objective Function)

Determine the constraint equation.

Since the volume has to hold 1000 ml, the constraint equation is the volume of the cylinder. If the volume of a cylinder is (V)=πr2(h), then the constraint equation is:

1000 =πr2 h (constraint equation)

Let’s isolate ‘h’ from the constraint equation:

Now, we will get the value of ‘h’ from the constraint equation.

h = 1000/πr2

Now, put this value of ‘h’ into the objective function, which gives us our combined objective function of the surface area of the cylinder.

S = 2πrh+2πr2

S = 2πr (1000/πr2) + 2πr2

S = 2000/r + 2πr2 or S = 2000r-1 + 2πr2

Perform the first and second derivative test on the combined objective function and test the critical points.

Take the derivative of the combined surface area function by using the power rule:

S = 2000r-1 + 2πr2

S’ = -2000/r2 + 4πr or -2000r-2 + 4πr

Remember that this equation will be valid only if the r > 0, if the r is 0, then the function will be undefined.

Set S’ equal to zero to find the critical numbers:

0 = -2000/r2 + 4πr

Solving for r,

r = 5.419

r = 5.419 is our only critical number.

Let’s take the second derivative of the surface area function:

S’’ = 4000r-3 + 4π

Substitute r = 5.419 into S’’.

S’’ = 4000r-3 + 4π

S’’ = 4000(5.419)-3 + 4π

S’’ = 37.07

Since S’’ is positive, the surface area function S is concave up, which means that at r = 5.419, there is a local minimum.

So, we will put the value of r = 5.419 into the constraint equation:

h = 1000/π(5.419)2

h = 10.839

Look back at the question to make sure you answered what was asked.

The dimensions that will minimize the cost of the cylinder are radius ‘r’ = 5.419 cm3 and height ‘h’ = 10.839 cm3.

Example 3:

The company you own has a large supply of 10 inch by 15inch rectangular pieces of tin, and you decide to make them into boxes by cutting a square from each corner and folding up the sides. How much should you cut from each corner, so the resulting box has the greatest volume?

Solution:

Translate the word problem into a picture. This will help you to visualize any shapes and write the appropriate dimensions.

A picture is provided below: Determine the objective function. Look for words indicating the largest or smallest value of something.

Since we are maximizing the volume, our objective function will be the volume:

The equation for the volume of a box is V = length * width * height

V = l*w*h

Determine the constraint equation.

In this case, we need to turn the length, width, and height as terms of x. Looking at the picture above, the length will be 15 – 2x, the width is 8 – 2x, and the height of the box is just x.

Therefore, the volume can be written as V(x) = (15 – 2x)(8 – 2x)(x) = 4x3-46x2+120x

Let’s envision the box being folded up to see the final product: Therefore, our combined objective function is V(x) = 4x3-46x2+120x

Perform the first and second derivative test on the combined objective function and test the critical points.

Take the derivative of the combined volume function by using the power rule:

V(x) = 4x3-46x2+120x

V ‘(x) = 12x2-92x+120

Set V’(x) equal to zero to find the critical numbers:

V ‘(x) = 12x2-92x+120

0 = 12x2-92x+120

Let’s use the quadratic formula to find the solution to 0 = 12x2-92x+120: Our critical points are x = 5/3 or x = 6.

Let’s find the second derivative of the volume function:

V ‘'(x) = 24x-92

Substituting the critical points into V’’(x),

V ‘'(5/3) = 24(5/3)-92 = -52 (concave down)

V ‘'(6) = 24(6)-92 = 52 (concave up)

V’’(x) is concave down for x = 5/3, then V(x) has a maximum at x = 5/3.

So, we will put the value of x = 5/3 into the constraint equation:

V(5/3) = 4(5/3)3-46x(5/3)2+120(5/3) = 90.74 cubic inches.

Look back at the question to make sure you answered what was asked.

The maximum volume of the box occurs when a 5/3 inch by 5/3 inch square is removed from each corner, and resulting box is 5/3 inches high, 8 – 2(5/3) = 14/3 inches wide, and 15 – 2(5/3) = 35/3 inches long.

Below are some videos that show you how to solve more optimization problems:

Optimization: Minimized the Surface are of an Open Top Box

Ex 3: Max / Min Application Problem – Rectangle in a Semicircle

Ex 2: Max / Min Application Problem – Max Volume of a Box

Ex: Optimization – Minimize the Cost to Make a Can with a Fixed Volume