Question

The shipping company BigBlue has a size requirement for rectangular packages: the sum of the three dimensions cannot exceed 255 inches. If one end of the package is a square with sides of x inches, and the other dimension is y inches, what are the dimensions that will maximize the volume?

x =

Answer 1

Your restriction for the volume is\[255 = x + x + y = 2x + y\]The volume is\[V = x^2 y\]You can either maximize x or y and then solve for the other. In this case, lets just maximize x.\[255 = 2x + y \rightarrow y = 255 - 2x\]\[y = 255 - 2x \rightarrow V = x^2 y = x^2 (255 - 2x)\]

Answer 2

\[V = 255x^2 - 2x^3\]Its max will occur when the derivative = 0.\[V' = 510 x - 6x^2 = 0\]\[x(510 - 6x) = 0\]\[x = 0, x = \frac {510}{6} = 85\]Obviously x can't = 0, so the max volume will occur when x = 85 inches. Plug that back in to get the y value\[x = 85 \rightarrow y = 255 - 2x = 255 - 170 = 85\]So when the volume will be maximized with the restriction, when all three dimensions are the same, x = y = 85 in.

Answer 3

so wait x and y are the same?

Answer 4

Yup, they're both 85 in.

Answer 5

how could i find the max volume w. what you found?

Answer 6

Did you learn about local max or local min in class?

Answer 7

we have learned about relative max and min

Answer 8

Yeah, its the same concept, just applied to the real world. After you get the volume equation, find its x coordinate that will make it a relative max.

Answer 9

so plug in 85 into the volume equatio and solve for x?

Answer 10

Not quite, after you solve for everything your answer is 85 inches...
I can't explain fully, but look at example 2 in this link to help you get an idea.
http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx

Answer 11

ok thanks!

Answer 12

Alright, no problem, I hope you understand it :)