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#
length #= 3x = 5.646 ft#
height #= y = 4.705 ft#

Explanation:

STEP 1: Find the Surface Area equation:
top and bottom = #3x# by #x# => Area = #3x^2#
sides = #x# by #y #and #3x# by #y# => Area = #xy# and #3xy#
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 + 8xy#

STEP 2: Find the volume equation (Draw a picture):
Set width equal to #x#, and height equal to #y#

Volume = #x# by #3x# by #y = 3x^2y#

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)#

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

#SA = 6x^2 + 8xy = 6x^2 + 8x((50)/(3x^2)) = 6x^2 + (400)/(3x)#

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

multiply the top/bottom #SA# by 10, because that is the cost to make the top and bottom, and the sides #SA# 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^-1#

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 #x#values 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#