1. Home
  2. Calculus volume 1
  3. Applications of derivatives
  4. Applied optimization problems

4.7 Applied optimization problems  (Page 2/8)

<< Chapter < Page
  Calculus volume 1     Page 2 / 8
Chapter >> Page >

Determine the maximum area if we want to make the same rectangular garden as in [link] , but we have 200 ft of fencing.

The maximum area is 5000 ft 2 .

Got questions? Get instant answers now!

Now let’s look at a general strategy for solving optimization problems similar to [link] .

Problem-solving strategy: solving optimization problems

  1. Introduce all variables. If applicable, draw a figure and label all variables.
  2. Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).
  3. Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.
  4. Write any equations relating the independent variables in the formula from step 3 . Use these equations to write the quantity to be maximized or minimized as a function of one variable.
  5. Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.
  6. Locate the maximum or minimum value of the function from step 4 . This step typically involves looking for critical points and evaluating a function at endpoints.

Now let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.

Maximizing the volume of a box

An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?

Step 1: Let x be the side length of the square to be removed from each corner ( [link] ). Then, the remaining four flaps can be folded up to form an open-top box. Let V be the volume of the resulting box.

There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 – 2x in, and 36 – 2x in.
A square with side length x inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.

Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V .

Step 3: As mentioned in step 2 , are trying to maximize the volume of a box. The volume of a box is V = L · W · H , where L , W , and H are the length, width, and height, respectively.

Step 4: From [link] , we see that the height of the box is x inches, the length is 36 − 2 x inches, and the width is 24 − 2 x inches. Therefore, the volume of the box is

V ( x ) = ( 36 − 2 x ) ( 24 − 2 x ) x = 4 x 3 − 120 x 2 + 864 x .

Step 5: To determine the domain of consideration, let’s examine [link] . Certainly, we need x > 0 . Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for x over the open interval ( 0 , 12 ) . Since V is a continuous function over the closed interval [ 0 , 12 ] , we know V will have an absolute maximum over the closed interval. Therefore, we consider V over the closed interval [ 0 , 12 ] and check whether the absolute maximum occurs at an interior point.

Step 6: Since V ( x ) is a continuous function over the closed, bounded interval [ 0 , 12 ] , V must have an absolute maximum (and an absolute minimum). Since V ( x ) = 0 at the endpoints and V ( x ) > 0 for 0 < x < 12 , the maximum must occur at a critical point. The derivative is

V ′ ( x ) = 12 x 2 − 240 x + 864 .

To find the critical points, we need to solve the equation

12 x 2 − 240 x + 864 = 0 .

Dividing both sides of this equation by 12 , the problem simplifies to solving the equation

x 2 − 20 x + 72 = 0 .

Using the quadratic formula, we find that the critical points are

x = 20 ± ( −20 ) 2 − 4 ( 1 ) ( 72 ) 2 = 20 ± 112 2 = 20 ± 4 7 2 = 10 ± 2 7 .

Since 10 + 2 7 is not in the domain of consideration, the only critical point we need to consider is 10 − 2 7 . Therefore, the volume is maximized if we let x = 10 − 2 7 in . The maximum volume is V ( 10 − 2 7 ) = 640 + 448 7 ≈ 1825 in . 3 as shown in the following graph.

The function V(x) = 4x3 – 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads “Maximum volume is approximately 1825 cubic inches when x ≈ 4.7 inches.”
Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.
Got questions? Get instant answers now!
<< Chapter < Page Page > Chapter >>
Practice Key Terms 1

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask