SOLUTION: please help having trouble trying to figure out this problem... A cardboard box has a square base, with each edge of the base having length x inches. The total length of all 12

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Question 413802: please help having trouble trying to figure out this problem...
A cardboard box has a square base, with each edge of the base having length x inches. The total length of all 12 edges of the box is 200 in.

a. Express the volume V of the box as a function of x .
V(x) =
b. What is the domain of V ? [Hint: Assume that a zero volume is not allowed.]

c. Offline draw a graph of the function V and use it to find the maximum volume for such a box
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A cardboard box has a square base, with each edge of the base having length x inches.
The total length of all 12 edges of the box is 200 in.
:
Let the dimensions of the box = x by x by h
then: 8 edges = x, and 4 edges = h
therefore:
8x + 4h = 200
simplify, divide by 4
2x + h = 50
h = (50-2x); we can use this for substitution
:
a. Express the volume V of the box as a function of x.
V = x^2 * h
replace h with (50-2x)
V(x) = x^2*(50-2x)
V(x) = -2x^3 + 50x^2
:
b. What is the domain of V ? [Hint: Assume that a zero volume is not allowed.]
domain |x| x > 0; (you can't have negative dimensions)
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I made a mistake here, should be |x| x > 0, x < 25, as you can see by the graph)
As you can see when x=25
V(25)=-2(25^3) + 50(25^2
V(25)= -2(15625) + 50(625)
V(25)= -31250 + 31250, we end up with 0 volume again
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:
c. Offline draw a graph of the function V and use it to find the maximum volume for such a box
plot the above equation

From this you can see that max volume occurs when x ~ 17 inches
my Ti83, gave the exact max of x = 16
:
:
We can confirm this using 16.67 inches
Find the height: 50 - 2(16.67) = 16.67; (so max area is perfect cube)
8(16.67) + 4(16.67) ~ 200
:
:
Did I explain this well enough for you to understand what went on here?