- #1
illjazz
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Homework Statement
If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
Homework Equations
- Differentiation rules
- Magic?
The Attempt at a Solution
First, I drew a square. I drew little squares in the corners of the big square. The sides of the little squares I labeled y. Each side of the big square has a length of [tex]\sqrt{1200}[/tex]. The length of the sides of the box that is to be built is [tex]\sqrt{1200} - 2y[/tex] and is labeled x.
The volume of the box is given by
[tex]V=yx^2[/tex]
We know [tex]x=\sqrt{1200}-2y=20\sqrt{3}-2y[/tex]
So we get a function
[tex]V(y)=y(20\sqrt{3}-2y)^2[/tex]
The next step, I presume, is to take the derivative of this function. Then we have
[tex]V'(y)=4(3y^2-40y\sqrt{3}+300)[/tex]
As a first step in finding a maximum for this function, I set V' equal to zero:
[tex]4(3y^2-40y\sqrt{3}+300)=0[/tex]
I get
[tex]y=0[/tex] or [tex]y=10\sqrt{3}[/tex] by asking my calculator.
(Trying to calculate the "zeroes" for this equation by hand ends in disaster..
Here's my attempt:
[tex]12y^2-160y\sqrt{3}+1200=0[/tex]
[tex]12y^2-160y\sqrt{3}=-1200[/tex]
[tex]3y^2-40y\sqrt{3}=-300[/tex]
[tex]y^2-\frac{40\sqrt{3}}{3}y=-100[/tex]
Eh!?
How would I go from here to get the zeroes? That's just a side question..)
So, now I can make a number line, or a "wiggle graph":
[tex]V'(-1)=40(40\sqrt{3}+303)[/tex], which is positive.
[tex]V'(1)=-40(40\sqrt{3}+303)[/tex], which is positive.
[tex]V'(10\sqrt{3}+1)=40(20\sqrt{3}+3)[/tex], which is positive...
Code:
---+-----------+--------------------+------
---------(0)---------(10\sqrt{3})-----------
Hmm.. they're not supposed to be all positive. How am I supposed to find a maximum if V' never goes from positive to negative? There is obviously something wrong here..