Question 66207
an open top box is to be made so that its width is 4 ft and its volume is 40 ft^3. the base of the box costs $4/ft^2and the sides cost $2/ft^2.
:
A) express the cost of the box as a function of its length (L) and height (h).
:
Cost = base cost + 4 sides cost (2 of one size and 2 of another size)
:
Cost = $4(4*L) + 2[$2(4*h)] + 2[$2(L*h)]
Cost = 16L + 16h + 4L*h
:
:
B) find a relationship between (L) and (h)
:
Using the vol equation: 
L*w*h = 40
L*4*h = 40
4L*h =40
L*h = 40/4
L*h = 10
:
L = 10/h
and
h = 10/L
:
:
C) express the cost as a function of (h) only
:
Using the cost equation: Cost = 16L + 16h + 4L*h, substitute (10/h) for L:
Cost = 16(10/h) + 16h + 4[(10/h)*h)
Cost = (160/h) + 16h + 40; note that the h's cancel
:
:
D) give the domain of the cost function
(0,+infinity)
:
:
E) using a graphing calculator or computer to approximate the dimensions of the box having the least cost.
Enter:
 y = (160/x) + 16x + 40;  x represents the height and y represents the cost.
Using the minimum feature I got x = 3.162, (height); y = 141.19, (min cost)
:
It's interesting to note that the min cost occurs when both the height and the
length = 3.162, so two of the sides are square. (L = 10/3.162 = 3.162)
:
Dimensions of the box: 3.162 x 4 x 3.162 = 39.9929, (close enough to 40 cu ft)
:
Hope this helped you, it doesn't seem the the domain could increase forever
and practically it couldn't, but no algebra reason for it not to?? Ask your
instructor about that.