## tmagloire1 one year ago A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.)

1. SolomonZelman

calc I prob optimization, correct?

2. SolomonZelman

|dw:1437076558138:dw|

3. tmagloire1

Yep optimization

4. tmagloire1

I got 2029.95 but that isn't correct

5. SolomonZelman

|dw:1437076685341:dw|

6. tmagloire1

Yep that's how i set it up

7. SolomonZelman

do you know why I subtracted 2xfrom 20 and 50?

8. SolomonZelman

So you do know why we subtract 2x from 20 and 50, right/

9. tmagloire1

Yes i understand

10. SolomonZelman

So, now, we will create the function for volume. Find critical numbers Find the maximum of it (with limits we will later provide/make)

11. SolomonZelman

The volume would be height•lenght•width

12. tmagloire1

I got 4.468

13. SolomonZelman

and in this case v(x)=2x•(20-2x)•(50-2x)

14. SolomonZelman

where V stands for volume

15. tmagloire1

Why would it be 2x instead of just x

16. SolomonZelman

u need v(x)

17. SolomonZelman

Why would it be 2x instead of just x, because you bend a cut-out of length x from both sides, don't you?

18. SolomonZelman

|dw:1437076945620:dw|

19. tmagloire1

ohh okay

20. SolomonZelman

so the side 20 and side 50 is decreased by 2x. Got that, good:)

21. tmagloire1

So now take the derivative of the function we got?

22. SolomonZelman

now we will do a regular process once we got the function V(x)=2x•(20-2x)•(50-2x)

23. SolomonZelman

I would perhaps simplify it a little instead of doing a product twice, but you can find the derivative as you wish:)

24. SolomonZelman

25. SolomonZelman

V(x)=x•(20-2x)•(50-2x) ^ ^ just x, not 2x.

26. SolomonZelman

but 20-2x and 50-2x are correct measures

27. tmagloire1

Okay so take the derivative of that then?

28. SolomonZelman

V(x)=x•(20-2x)•(50-2x) V(x)=x•(1000-40x-100x+4x²) V(x)=x•(1000-140x+4x²) I would do derivative, product now

29. SolomonZelman

V$$\rm ~'$$(x) = 1•(1000-140x+4x²)+x•(0-140+8x)

30. tmagloire1

So it will be 12x^2-280x-1000?

31. SolomonZelman

I would use wolfram, actually, because I am the laziest person on this planet.

32. tmagloire1

Hahaha i just used Wolfram for it because i didn't feel like taking the derivative myself xD

33. SolomonZelman

V'(x)=4(3x²-70x+250) by wolfram.... I did it like this because wolfram can give a handy version faster

34. SolomonZelman

set V'(x)=0 to find your critical number

35. SolomonZelman

numbers*

36. SolomonZelman

remember tho, that 0>x>10 or else your volume is going to be =0 or negative.

37. tmagloire1

I got 11.66 :/

38. SolomonZelman

nope

39. SolomonZelman

that wouldn't make sense, because you get a negative volume if you plug it in our volume function.

40. SolomonZelman

As again 0<x<10

41. SolomonZelman

find the roots of 4(3x²-70x+250)=0

42. tmagloire1

26.48 and -3.146 which still doesn't make sense right?

43. SolomonZelman

And I will tell you will get the ±, and only take the minu one, because the one with the + is greater than 10

44. SolomonZelman

what crtical number(s) did you get?

45. tmagloire1

I am getting -60 now :// sos

46. SolomonZelman

|dw:1437077863587:dw|

47. SolomonZelman

take the -, because + will get x>10

48. SolomonZelman

35-5√19 ≈ 4.4

49. tmagloire1

Oh I must have been doing some crazy math error when taking the crit #

50. tmagloire1

so take 4.4 and use it in the equation at the beginning?

51. SolomonZelman

yes,

52. tmagloire1

i got -2000 ._.

53. SolomonZelman

you did something wrong again-:(

54. tmagloire1

2030.3*

55. SolomonZelman

Yup, that is right

56. SolomonZelman

2030.336

57. tmagloire1

Oh okay! Thank you so much of all the help. I think I understand it now!!

58. SolomonZelman

Yes.... overview. 1. We created a function V(x)=x • (20-2x) • (50-2x) 2. Constraint x>0, because otherwise volume is negative or 0 (based on the first, x, component of V(x)). x<10, because otherwise (based on the 20-2x part), the volume would be 0 or negative. 3. We differentiated the function to get 2 critical numbers (that is two solutions of V'(x)=0) NOTE: Critical numbers (usually) are also values for which v'(x) is undefined, if defined for initial function v(x) - (but no such values in this case). Also, critical numbers usually can be closed interval boundaries, but our interval is open (we didn't include 0 or 10), so that is not our case. So in our case only solutions to v(x)=0 are critical numbers. 4. We excluded the (35+5√19)/3 solution based on our constraint (because it is greater than 10), and rather used the only solution remaining, i.e. (35-5√19)/3. 5. We got the answr

59. SolomonZelman

yw