## anonymous 5 years ago An open box is being constructed whose base length is 3 times the base width and whose volume is 50 cubic meters. If thematerials used to build the box cost $10 per square meter for the bottom and$6 per square meter for the sides, what are the dimensions for the least expensive box? Help?!?! I feel as if not enough information is provided. Yet again I could be wrong? How do I solve this problem?

1. anonymous

2. anonymous

this is an optimization problem, did you see that in class?

3. anonymous

Yes I do recognize that its an optimization problem, in which we have to alter our formulas.

4. anonymous

Help?

5. anonymous

6. anonymous

or set up

7. anonymous

8. anonymous

okay, so first lets take care of the constraint. i.e. the volume of this box is 50 cm^3

9. anonymous

yes

10. anonymous

how about $V=3x^2y$ and we know that $50=3x^2y$

11. anonymous

is that good?

12. anonymous

where did the 3x^2 come from?

13. anonymous

woah sorry i had to go fix a computer ill let ebbflo take over hes got a good flow going,

14. anonymous

** I meant 3x^2y

15. anonymous

the length is 3 time the width

16. anonymous

so 3x^2 is the area of the base and I just let y be the heigth of the box

17. anonymous

is that good?

18. anonymous

yes

19. anonymous

okay now we need the surface area formula for this box which since there is no top is $S=3x(x)+2(3x)y+2xy$

20. anonymous

Area = base * height so the base =3x and the height=x

21. anonymous

i am just adding up the areas of the sides of the box

22. anonymous

the base of the box has dimensions (3x)cm by (x)cm

23. anonymous

are you good with the surface area formula I presented?

24. anonymous

If so We can proceed to the cost function

25. anonymous

im sorry meters

26. anonymous

can you explain to me how you arrived to your Surface area

27. anonymous

since this box only has 5 sides we just add up the areas of those sides the area of the bottom is (3x)(x) the combined area of one of the pairs of equal sides is 2(3x)(y) and the combined area of the remaining pair of equal sides is 2(x)(y) then I just summed those up

28. anonymous

so that's where I got my formula for the surface area

29. anonymous

allright.

30. anonymous

so now ti get the cost function we just multiply 10 by the area of the bottom and 6 times each of the combined areas of the sides like so $C(x,y)=10(3x^2)+6(6xy)+6(3xy)$ collecting like terms we pretty it up to get $C(x,y)=30x^2+48xy$

31. anonymous

sorry I made a slight mistake in that formula

32. anonymous

the last one I wrote was correct

33. anonymous

$C(x,y)=10(3x^2)+6(6xy)+6(2xy)$ should have been the first one from which I got $C(x,y)=30x^2+48xy$

34. anonymous

now we are ready to write the cost function above as a function of x only by way of the constraint

35. anonymous

okay.

36. anonymous

so since $50=3x^2y$$y=\frac{50}{3x^2}$

37. anonymous

so $C(x)=30x^2+48x(\frac{50}{3x^2})$ and cleaning it up we get $C(x)=30x^2+\frac{800}{x}$

38. anonymous

if we are good, then now it is time to derive

39. anonymous

$C^\prime(x)=60x-\frac{800}{x^2}$ get a common denominator to make finding the critical points easier we get$C^\prime(x)=\frac{60x^3-800}{x^2}$

40. anonymous

since this is a box with known volume we know $x\neq 0$ so just solve $60x^3-800=0$ for x

41. anonymous

we get $x=2(\frac{5}{3})^\frac{1}{3}$

42. anonymous

this is the only value we get so many people quit there and just go with but you can confirm this is a minimum value via the second derivative test for max/min

43. anonymous

use that vale to get the other dimensions and you are done...

44. anonymous

thank you soo much

45. anonymous

glad it helps, this is easier to explain with the aid of a sketch

46. anonymous

I was just looking over the work and you told me to use my x value to figure out my other dimensions. Do i do that by taking the second derivative?

47. anonymous

no just find the height by plugging that x value into the constraint we solved for why and the other dimension is just 3 time that x value we got

Find more explanations on OpenStudy