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## Lesson 17: Optimization Problems

As businesses continue to compete, they seek to minimize costs and maximize profits. It is for this reason that optimization problems are so useful. A factory wants to produce at the output level that will maximize the profits while minimizing the cost of production. In a similar way, a farmer wants to plant the precise number of seeds that will minimize planting costs while maximizing his yield.

Below are a series of steps which should serve as guidelines to you when you go to solve an optimization problem.

A lifeguard at the beach has 400 m of rope to enclose an area at the beach for swimming. The beach will form one side of a rectangle and the rope will form the other 3 sides. Find the dimensions of the rectangle that will produce the maximum area assuming there are no restrictions on the rectangle. Then assume that the rope can only be let out 30 m into the water. What are the dimensions that produce the maximum area then?

In solving the problem we will go step by step over the guidelines given above. First we note that the question is asking for the maximum area to be enclosed by a rope, which is to create a rectangle with the beach. Next, we can define all the variables and include a diagram.

Below are a series of steps which should serve as guidelines to you when you go to solve an optimization problem.

- Identify what is being asked by the question
- Define all variables, drawing a diagram as needed.
- Identify the quantity to be optimized and write an equation.
- Define an independent variable and express all other variables in terms of the independent variable.
- Define a function in terms of the independent variable.
- State all restrictions on the independent variable.
- Differentiate the function.
- Calculate and classify all critical points.
- Answer the question given by the problem.

A lifeguard at the beach has 400 m of rope to enclose an area at the beach for swimming. The beach will form one side of a rectangle and the rope will form the other 3 sides. Find the dimensions of the rectangle that will produce the maximum area assuming there are no restrictions on the rectangle. Then assume that the rope can only be let out 30 m into the water. What are the dimensions that produce the maximum area then?

In solving the problem we will go step by step over the guidelines given above. First we note that the question is asking for the maximum area to be enclosed by a rope, which is to create a rectangle with the beach. Next, we can define all the variables and include a diagram.

Let A represent the area of the rectangle encompassed by rope.

Let L represent the length of the rectangle created.

Let W represent the width of rectangle created.

Let L represent the length of the rectangle created.

Let W represent the width of rectangle created.

Next we identify the quantity to be optimized and we write an equation. The quantity to be optimized is the Area of the rectangle, represented by the variable A.

The equation representing the Area of the enclosure is:

A=W×L

Next we determine that the independent variable is W, the width, because there are two parts to this problem. In the second part of the problem we will have a restriction on the width, so the width should be the independent variable.

2W+L=400

The next step is to express the function in terms of our independent variable. Note that we have to express one function in terms of W so that we can maximize this variable. To do this, isolate for the other variable, L and substitute into the other equation. This will give us one function in terms of W:

2W+L=400

L=400-2W Now substitute this into A=W×L.

A=W×(400-2W) Simplify.

A=400W-2W²

Here we are required to state all restrictions on the independent variable. For this part of the problem the only restrictions are based on the fact that the total length of rope must be less than 400 m. Obviously the minimum value is 0. The maximum value assumes that the length is 0. It can be computed by isolating W:

2W+L=400

2W+0=400

W=200

Therefore 0≤W≤200. Later on, when we solve the next part of the problem, we will resume at this step. For now, we must find the critical points of A by differentiating and solving for A'=0. Let us begin.

A=400W-2W²

A'=400-4W

0=400-4W

4W=400

W=100

We have a critical point at W=100. Solve for A so that we have the full co-ordinate.

2W+L=400

2(100)+L=400

200+L=400

L=200

A=W×L

A=100×200

A=20,000

The critical point is (100,20000). We can check whether this is a local maximum or a local minimum by taking the second derivative.

A=400W-2W²

A'=400-4W

A''=-4

Since the second derivative is -4 over all values, the critical point (100,20000) is a local maximum. As a result, the dimensions that allow for a maximum area of 20000m² are a width of 100m and a length of 200m. Next let us resume the problem assuming that the rope can only be let out 30 m into the water. We are looking for maximum dimensions over the interval 0<L<30. From earlier we remember that the derivative has one local maximum at W=100. We can infer that the function is increasing continuously until the local maximum and then decreasing continuously. Let us test the area for W=0 and W=200 to make sure:

2W+L=400

2(0)+L=400

L=400

A=W×L

A=0×400

A=0

2W+L=400

2(200)+L=400

L=0

A=W×L

A=200×0

A=0

Since the function is increasing continuously until the local maximum, the right endpoint will yield the maximum area. That is a width of 30 m will yield the maximum area. Based on this, we can calculate all dimensions.

2W+L=400

2(30)+L=400

60+L=400

L=340

A=W×L

A=30×340

A=10200

Therefore the dimensions that allow for the maximum area of 10,200 m², given the restriction, are a width of 30 m and a length of 340 m. Let's solve another problem involving volume.

A box must be manufactured from a certain plastic. The box must have a square base and a volume of 27 L. First find the dimensions that will minimize the plastic used for the box. Then, find the dimensions of the box that will minimize cost given that the plastic costs 0.3 ¢/cm², except for the plastic at the base which is thicker and costs 4 times as much.

The first part of the problem is asking us for the dimensions off plastic that will minimize the total plastic used up. First let us write out our variables and draw a diagram.

The equation representing the Area of the enclosure is:

A=W×L

Next we determine that the independent variable is W, the width, because there are two parts to this problem. In the second part of the problem we will have a restriction on the width, so the width should be the independent variable.

2W+L=400

The next step is to express the function in terms of our independent variable. Note that we have to express one function in terms of W so that we can maximize this variable. To do this, isolate for the other variable, L and substitute into the other equation. This will give us one function in terms of W:

2W+L=400

L=400-2W Now substitute this into A=W×L.

A=W×(400-2W) Simplify.

A=400W-2W²

Here we are required to state all restrictions on the independent variable. For this part of the problem the only restrictions are based on the fact that the total length of rope must be less than 400 m. Obviously the minimum value is 0. The maximum value assumes that the length is 0. It can be computed by isolating W:

2W+L=400

2W+0=400

W=200

Therefore 0≤W≤200. Later on, when we solve the next part of the problem, we will resume at this step. For now, we must find the critical points of A by differentiating and solving for A'=0. Let us begin.

A=400W-2W²

A'=400-4W

0=400-4W

4W=400

W=100

We have a critical point at W=100. Solve for A so that we have the full co-ordinate.

2W+L=400

2(100)+L=400

200+L=400

L=200

A=W×L

A=100×200

A=20,000

The critical point is (100,20000). We can check whether this is a local maximum or a local minimum by taking the second derivative.

A=400W-2W²

A'=400-4W

A''=-4

Since the second derivative is -4 over all values, the critical point (100,20000) is a local maximum. As a result, the dimensions that allow for a maximum area of 20000m² are a width of 100m and a length of 200m. Next let us resume the problem assuming that the rope can only be let out 30 m into the water. We are looking for maximum dimensions over the interval 0<L<30. From earlier we remember that the derivative has one local maximum at W=100. We can infer that the function is increasing continuously until the local maximum and then decreasing continuously. Let us test the area for W=0 and W=200 to make sure:

2W+L=400

2(0)+L=400

L=400

A=W×L

A=0×400

A=0

2W+L=400

2(200)+L=400

L=0

A=W×L

A=200×0

A=0

Since the function is increasing continuously until the local maximum, the right endpoint will yield the maximum area. That is a width of 30 m will yield the maximum area. Based on this, we can calculate all dimensions.

2W+L=400

2(30)+L=400

60+L=400

L=340

A=W×L

A=30×340

A=10200

Therefore the dimensions that allow for the maximum area of 10,200 m², given the restriction, are a width of 30 m and a length of 340 m. Let's solve another problem involving volume.

A box must be manufactured from a certain plastic. The box must have a square base and a volume of 27 L. First find the dimensions that will minimize the plastic used for the box. Then, find the dimensions of the box that will minimize cost given that the plastic costs 0.3 ¢/cm², except for the plastic at the base which is thicker and costs 4 times as much.

The first part of the problem is asking us for the dimensions off plastic that will minimize the total plastic used up. First let us write out our variables and draw a diagram.

Let SA represent the total area of plastic required to make the box, in cm².

Let V represent the volume of the box, in cm³.

Let x represent the side length of the square base, in cm.

Let h represent the height of the box, in cm.

Let V represent the volume of the box, in cm³.

Let x represent the side length of the square base, in cm.

Let h represent the height of the box, in cm.

The quantity to be minimized is the SA. We begin by noting that the total volume must be 27L, Note that 1mL=1cm³, so 27 L is equal to 27000cm³. Let us begin by writing the volume equation along with the surface area equation.

V=x²h

SA=2x²+4xh

Note that we want everything to be in terms of one variable, so that we can minimize the surface area. It makes sense to write out the SA equation in terms of x, so that we can minimize it. Therefore, rewrite SA in terms of the independent variable x. To achieve this, isolate h in the volume equation:

V=x²h

h=V/x²

h=27000/x²

Now that we have h in terms of x, substitute this into the equation for SA:

SA=2x²+4xh

SA=2x²+4x(27000/x²)

SA=2x²+108000((x)^(-1))

There are no restrictions on the values x can take on. Now take the derivative and solve for critical points.

SA=2x²+108000((x)^(-1))

SA'=4x+108000(-1)((x)^(-2))

SA'=4x³-108000/((x)^(2)) Rewrite as a fraction because it will be easier to take the zero in this way.

Now determine all critical points by letting SA' equal zero and solving.

SA'=4x³-108000/((x)^(2))

0=4x³-108000/((x)^(2)) Let the numerator equal zero and solve.

0=4x³-108000/((x)^(2))

0=4x³-108000

0=4x³-108000

4x³=108000

x³=27000

x=30

To classify x=30, we will verify that SA' changes signs and check if SA'' is positive or negative.

SA'=4x³-108000/((x)^(2))

SA'(25)=4(25)³-108000/((25)^(2)) Test a value that is less than x=30.

SA'(25)=62500-108000/(625)

SA'(25)=-72.8 The negative result denotes that the original function is decreasing.

SA'=4x³-108000/((x)^(2))

SA'(32)=4(32)³-108000/((32)^(2)) Test a value that is less than x=30.

SA'(32)=131072-108000/(1024)

SA'(32)=22.531 The positive result denotes that the original function is increasing.

SA'=4x³-108000/((x)^(2))

SA'=4x-108000((x)^(-2)) Prepare for differentiation by rewriting the x term so that there is no denominator.

SA''=4-(-2)108000((x)^(-3))

SA''=4+[216000/((x)^(3))]

Now that we have the second derivative, solve SA''(30) to determine whether we are dealing with a local maximum or minimum.

SA''(30)=4+[216000/((30)^(3))]

SA''(30)=4+[216000/27000]

SA''(30)=4+[8]

SA''(30)=12

Since the second derivative is positive, the curve is concave up and we have a minimum at x=30. Substitute this into the initial volume equation to derive the dimensions of the prism which utilizes the least plastic.

V=x²h

h=V/x²

h=27000/(30)²

h=27000/900

h=30

Therefore, the dimensions that minimize the use of plastic used are 30cm by 30 cm by 30cm. Consequently, a cube utilizes the least plastic. Let us move on to the second part of the problem. The problem is asking us what dimensions would cost the least, given the cost of plastic. All the variables are already defined but we need to write a new equation because we need to take the cost of the plastic into account this time. We introduce a new variable, cost, and rewrite our equations.

Let C represent the cost of producing the rectangular prism.

SA=2x²+4x(27000/x²)

SA=x²+x²+108000/x Note that we separate x² from x² because the cost of the top and bottom are different. This will appear shortly.

C=1.2x²+0.3x²+0.3(108000/x)

C=1.5x²+32400/x The only restriction on x is that x cannot be equal to zero.

Now we have an equation for the cost. Let us differentiate this equation to determine the minimum cost.

C=1.5x²+32400/x

C'=3x-32400/x²

C'=[3x³-32400]/x² Let C' equal to zero to solve for the critical number.

0=[3x³-32400]/x² Multiply both sides by x² so that the numerator is equal to zero.

0=3x³-32400

10800=x³

x=22.104

We will verify that this is a minimum point by taking the second derivative and checking to see if we get a positive number for this value of x. This is because a positive result for the second derivative denotes that the function is concave up.

C'=3x-32400/x²

C'=3x-32400(x^(-2))

C''=3-32400(-2)(x^(-3))

C''=3+64800(x^(-3))

C''=3+64800/(x^(3))

C''(22.104)=3+64800/((22.104)^(3))

C''(22.104)=3+64800/((22.104)^(3))

C''(22.104)=9

Since the second derivative is positive, x=22.104 is a local minimum. Note that the problem only asks for the dimensions that will minimize the cost, not for the actual cost. Substitute x=22.104 into the original formula for volume to determine the height of the prism that will minimize cost.

V=x²h

h=V/x²

h=27000/(22.104)²

h=55.261

Therefore, the dimensions of the rectangular prism that minimizes the cost are 55.26 by 22.10 cm by 22.10 cm. All in all, that's all there is to optimization problems. As long as you follow the step by step process you should have no problems. The first step is to identify what is being asked in the question. Then, all the variables must be defined, with a diagram if needed. Then the quantity to be optimized must be expressed in an equation. An independent variable should be defined so that all other variable are in terms of that variable and a function in terms of the variable should be derived. Then state any restrictions on the independent variable. Finally differentiate the function, calculate and classify the critical point or points and answer the question posed in the word problem.

V=x²h

SA=2x²+4xh

Note that we want everything to be in terms of one variable, so that we can minimize the surface area. It makes sense to write out the SA equation in terms of x, so that we can minimize it. Therefore, rewrite SA in terms of the independent variable x. To achieve this, isolate h in the volume equation:

V=x²h

h=V/x²

h=27000/x²

Now that we have h in terms of x, substitute this into the equation for SA:

SA=2x²+4xh

SA=2x²+4x(27000/x²)

SA=2x²+108000((x)^(-1))

There are no restrictions on the values x can take on. Now take the derivative and solve for critical points.

SA=2x²+108000((x)^(-1))

SA'=4x+108000(-1)((x)^(-2))

SA'=4x³-108000/((x)^(2)) Rewrite as a fraction because it will be easier to take the zero in this way.

Now determine all critical points by letting SA' equal zero and solving.

SA'=4x³-108000/((x)^(2))

0=4x³-108000/((x)^(2)) Let the numerator equal zero and solve.

0=4x³-108000/((x)^(2))

0=4x³-108000

0=4x³-108000

4x³=108000

x³=27000

x=30

To classify x=30, we will verify that SA' changes signs and check if SA'' is positive or negative.

SA'=4x³-108000/((x)^(2))

SA'(25)=4(25)³-108000/((25)^(2)) Test a value that is less than x=30.

SA'(25)=62500-108000/(625)

SA'(25)=-72.8 The negative result denotes that the original function is decreasing.

SA'=4x³-108000/((x)^(2))

SA'(32)=4(32)³-108000/((32)^(2)) Test a value that is less than x=30.

SA'(32)=131072-108000/(1024)

SA'(32)=22.531 The positive result denotes that the original function is increasing.

SA'=4x³-108000/((x)^(2))

SA'=4x-108000((x)^(-2)) Prepare for differentiation by rewriting the x term so that there is no denominator.

SA''=4-(-2)108000((x)^(-3))

SA''=4+[216000/((x)^(3))]

Now that we have the second derivative, solve SA''(30) to determine whether we are dealing with a local maximum or minimum.

SA''(30)=4+[216000/((30)^(3))]

SA''(30)=4+[216000/27000]

SA''(30)=4+[8]

SA''(30)=12

Since the second derivative is positive, the curve is concave up and we have a minimum at x=30. Substitute this into the initial volume equation to derive the dimensions of the prism which utilizes the least plastic.

V=x²h

h=V/x²

h=27000/(30)²

h=27000/900

h=30

Therefore, the dimensions that minimize the use of plastic used are 30cm by 30 cm by 30cm. Consequently, a cube utilizes the least plastic. Let us move on to the second part of the problem. The problem is asking us what dimensions would cost the least, given the cost of plastic. All the variables are already defined but we need to write a new equation because we need to take the cost of the plastic into account this time. We introduce a new variable, cost, and rewrite our equations.

Let C represent the cost of producing the rectangular prism.

SA=2x²+4x(27000/x²)

SA=x²+x²+108000/x Note that we separate x² from x² because the cost of the top and bottom are different. This will appear shortly.

C=1.2x²+0.3x²+0.3(108000/x)

C=1.5x²+32400/x The only restriction on x is that x cannot be equal to zero.

Now we have an equation for the cost. Let us differentiate this equation to determine the minimum cost.

C=1.5x²+32400/x

C'=3x-32400/x²

C'=[3x³-32400]/x² Let C' equal to zero to solve for the critical number.

0=[3x³-32400]/x² Multiply both sides by x² so that the numerator is equal to zero.

0=3x³-32400

10800=x³

x=22.104

We will verify that this is a minimum point by taking the second derivative and checking to see if we get a positive number for this value of x. This is because a positive result for the second derivative denotes that the function is concave up.

C'=3x-32400/x²

C'=3x-32400(x^(-2))

C''=3-32400(-2)(x^(-3))

C''=3+64800(x^(-3))

C''=3+64800/(x^(3))

C''(22.104)=3+64800/((22.104)^(3))

C''(22.104)=3+64800/((22.104)^(3))

C''(22.104)=9

Since the second derivative is positive, x=22.104 is a local minimum. Note that the problem only asks for the dimensions that will minimize the cost, not for the actual cost. Substitute x=22.104 into the original formula for volume to determine the height of the prism that will minimize cost.

V=x²h

h=V/x²

h=27000/(22.104)²

h=55.261

Therefore, the dimensions of the rectangular prism that minimizes the cost are 55.26 by 22.10 cm by 22.10 cm. All in all, that's all there is to optimization problems. As long as you follow the step by step process you should have no problems. The first step is to identify what is being asked in the question. Then, all the variables must be defined, with a diagram if needed. Then the quantity to be optimized must be expressed in an equation. An independent variable should be defined so that all other variable are in terms of that variable and a function in terms of the variable should be derived. Then state any restrictions on the independent variable. Finally differentiate the function, calculate and classify the critical point or points and answer the question posed in the word problem.

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