Construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. Volume of the box is 50 ft3, what are the dimensions that will MINIMIZE the Cost?

1 Answer
Dec 6, 2017

width = x = 1.882 ft=x=1.882ft
length = 3x = 5.646 ft=3x=5.646ft
height = y = 4.705 ft=y=4.705ft

Explanation:

STEP 1: Find the Surface Area equation:
top and bottom = 3x3x by xx => Area = 3x^23x2
sides = xx by y yand 3x3x by yy => Area = xyxy and 3xy3xy
since we have two top/bottom, and two of each type of sides, add them up:

SA = 2( 3x^2 ) + 2(xy) + 2(3xy) = 6x^2 + 2xy + 6xy = 6x^2 + 8xySA=2(3x2)+2(xy)+2(3xy)=6x2+2xy+6xy=6x2+8xy

STEP 2: Find the volume equation (Draw a picture):
Set width equal to xx, and height equal to yy

Volume = xx by 3x3x by y = 3x^2yy=3x2y

STEP 3: To get our Surface Area equation all in terms of one variable, use what we know about Volume to find the relationship between x & y (solve for y in terms of x):

V = 50 = 3x^2y => y = (50)/(3x^2)V=50=3x2yy=503x2

STEP 4: Now that we know y=50/(3x^2)y=503x2, we can get the SASA equation in terms of one variable:

SA = 6x^2 + 8xy = 6x^2 + 8x((50)/(3x^2)) = 6x^2 + (400)/(3x)SA=6x2+8xy=6x2+8x(503x2)=6x2+4003x

STEP 5: Find the Cost equation, which we will optimize to find our answer:

multiply the top/bottom SASA by 10, because that is the cost to make the top and bottom, and the sides SASA by 6, then bring everything to the numerator (easiest way to differentiate later):

C = 10(6x^2) + 6((400)/(3x)) = 60x^2 + (800)/(x) = 60x^2 + 800x^-1C=10(6x2)+6(4003x)=60x2+800x=60x2+800x1

STEP 6: Differentiate Cost to find the minimum when C'=0 or C'DNE:

C' = 120x - 800x^-2

STEP 7: Multiply C' by (x^2)/(x^2) to get a form easier to find C' = 0 and C' DNE (We can do this because (x^2)/(x^2) = 1, and we can multiple anything by 1 and not change it):

C' = (120x - 800x^-2) ((x^2)/(x^2)) = (120x^3 - 800)/(x^2)

STEP 8: Find C'=0 or C'DNE:
C' = 0 (Use numerator): 120x^3 - 800 = 0 => 120x^3 = 800 => x^3 = (20)/(3) => x = root(3)((20)/(3)) ~~ 1.882
C' DNE (Use denominator): x^2=0 => x=0 (We can rule out this answer since x=0 does not make sense, we wouldn't have a box at all.)

STEP 9: Check and make sure 1.882 is a minimum by plugging in any value above and below 1.882 into our C' equation (lets use 1 and 2):

C'(1) = (120(1)^3 - 800)/((1)^2) = -680, which is a NEGATIVE
C'(2) = (120(2)^3 - 800)/((2)^2) = 40, which is POSITIVE

...So, since all xvalues of C' less than 1.882 are negative, and all x values of C' greater than 1.882 are positive, we can see there is a local minimum @ x=1.882. And since there are no endpoints, because we need a value/measurement greater than 0, 1.882 ft is our ABSOLUTE MINIMUM.

STEP 10: Now that we have our absolute minimum Cost @ x=1.882, plug into Volume equation to find y:

y = (50)/(3(1.882)^2) ~~ 4.706 ft

STEP 11 (Last Step!!): Find the Cost:

C(1.882) = 60(1.882)^2 + 800(1.882)^-1 ~~ $637.60