Maximizing volume of a Square-Based Pyramid?

[sidnygte]

Homework Statement

I am given an 8in x 8 in piece of paper and I need to build a pyramid with a square base with the greatest volume.

Homework Equations

V = 1/3 * x^2 * h

b = base length
h = height length

I'm trying to get everything in terms of x, which is the length of the base of the square I need to make.

The slant length is l

w(length of the entire slant) = 2l + xhttp://i39.tinypic.com/2me4jsw.jpg

Oh and l = h^2 + x^2/2. height squared + one half x^2.

The Attempt at a Solution

This is an optimization problem, so I put everything in terms of one variable, and then get the derivative
I just don't know if this is the derivative.

http://i44.tinypic.com/2rh07f5.jpg

 

[ehild]

Hi Sydneygte,

The slant length is l

w(length of the entire slant) = 2l + x


http://i39.tinypic.com/2me4jsw.jpg

Oh and l = h^2 + x^2/2. height squared + one half x^2.

You made a little mistake: It should be (x/2)^2 instead of x^2/2.
Also, type in the expression of the volume in terms of x and its derivative. You picture is not clear.

ehild

 

[sidnygte]

ohh that changes it a bit

so h^2 = l^2 - (x/2)^2

so you can square root everything so

h = l - (x/2)? that doesn't seem right

 

[SammyS]

ohh that changes it a bit

so h^2 = l^2 - (x/2)^2

so you can square root everything so

h = l - (x/2)? that doesn't seem right

If square roots worked like that, Pythagoras would have been out of business long ago.

Square roots don't work like that.

After all,

9 = 25 - 16 .​

But:

3 ≠ 5 - 4 .​

 

[sidnygte]

so what would the equation look like if you were trying to get H to the power of one?

h = sqrt(l^2 - .5x^2)?

 

[SammyS]

so what would the equation look like if you were trying to get H to the power of one?

h = sqrt(l^2 - .5x^2)?

It would be:

h = sqrt( l² - (0.5x)² )​

or equivalently

h = sqrt( l² - (0.25)x² )​

(That is a lower case L, not a number 1.)

 

[sidnygte]

so then the function in terms of x would be

V = 1/3X^2 * sqrt( sqrt(128)-x)^2 - (0.5x)^2) ?

that seems confusing haha. has would I simplify the double square root? Since it is being squared in the inside can I make it into

V = 1/3X^2 * sqrt(128 - X^2 - 0.25X^2)?

 

[SammyS]

You can't do that either .

(a-b)² ≠ a² - b² .

But the following is true:

(a-b)² = a² -2ab + b²

You will still have a square root if you "FOIL" ( sqrt(128)-x)².

BTW, √(128) = (8)√(2), and either way you write this, it's just a number, there's no variable involved.

Also , you are missing a parenthesis in " sqrt( sqrt(128)-x)^2 - (0.5x)^2) ".

It should be: sqrt( ( sqrt(128)-x)^2 - (0.5x)^2) .