66.9k views
5 votes
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 13 in. by 8 in.

1 Answer

5 votes
1) To make a rectangular box you need to cut squares from the four corners of the rectangular sheet.

2) Call x the length of the sides of the squares cut off.

3) The base of the box will have dimensions: (13 - 2x) and (8 - 2x)

4) The height of the box will be x

5) The volume of the box will be the area of the base times the height:

Volume = (13 - 2x)(8 -2x)x = (4x^2 - 42x + 104)x = 4x^3 - 42x^2 + 104x

6) The maximum volume is calculated by finding the point where the derivative of the volume is zero =>

d (volume) / dx = 12x^2 - 84x + 104 = 0

7) Solve the quadratic equation 12x^2 - 84x + 104 = 0

=> 4(3x^2 - 21x + 26) = 0

=> 3x^2 - 21x + 26 = 0

=> 3 (x^2 - 7x) + 26 = 0

=> 3 [(x - 7/2)^2 - (7/2)^2] + 26 = 0

=> 3 (x - 7/2)^2 - 3* 49/4 + 26 = 0

=> 3 (x - 7/2)^2 = 3*49/4 - 26

=> (x -7/2)^2 = (49/4 - 26/3)

=> x = 7/2 +/- √(49/4 - 26/3)

x = 7/2 + √3.583 and x = 7/2 - √3.583

x = 5.393 and x = 1.607

=> Volume =

1) 4(5.393)^3 - 42(5.393)^2 + 104(5.393) = -33.26 ---> it does not have physical meaning

2) 4(1.607)^3 - 42(1.607)^2 + 104(1.607) = 75.27 ---> this is the answer

Answer: 75.27 in^3



User FBente
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories