A box is to be made out of a 12 by 20 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L , width W , and height H of the resulting box that maximizes the volume.

Box Dimensions:

L  = 15.15 ul

W = 7.15 ul

h = x = 2.43 ul

V(max) =    263.22 cu

Step-by-step explanation:

We call x the length of the square to be cut in the corners then:

Are of the base of the box is:

(20 - 2*x)  is the future length of the box and

(12 - 2*x) will be the width

The heigh is x  then the volume of the box is:

V = ( 20 - 2*x )* ( 12 - 2*x ) * h

And the volume as a function of x is:

V(x) = ( 20 - 2*x) * ( 12 - 2*x ) * x     or   V(x) = (240 -40*x -24*x + 4*x²) * x

V(x) =  240*x - 64*x² + 4*x³

Taking derivatives on both sides of the equation we get:

V´(x) = 240 - 128*x + 12*x²

V´(x)  =  0              240 - 128*x + 12*x²  = 0    or     60 - 32*x + 3*x²

3*x²  -  32*x  +  60  = 0

Solving:

x₁,₂  =  32  ± √ (32)² - 4*3*60 ]/ 2*3

x₁,₂  =  32  ± √ 1024 - 720 )/6

x₁,₂  = ( 32  ± √ 304 )/6

x₁,₂  = ( 32  ± 17.44 )/6

x₁  =  8.23      ( we dismiss this solution because is not feasible 2*x > 12

x₂ = 2.43  u.l ( units of length)

Then

L  =  20 - 2*x      L  = 20  - 4.85     L  = 15.15 ul

W =  12 - 2*x      W = 12 - 4.85      W = 7.15 ul

h  =  2.43 ul

V = 2.43*7.15*15.15  cubic units

V =  263.22 cu

To see if when x = 2.43   function V has a maximum we go to the second derivative

V´´(x)  = - 128  + (24)*2.43

V´´(x) = - 69.68      as        V´´(x)  < 0   then we have a maximum for V(x) in the point x = 2.43

the f

last graph i believe

step-by-step explanation:

worked it out i believe

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068