Question

A pyramid has a square base of side x (in cm), and a height h (in cm) and its volume V is given by the formula V = 1/3x^2h. If the sum of the length of the base side and the height is 9, find the max volume. Give the dimensions of maximum volume.

Answer 1

Use differentiation to find the first and second derivative.

Answer 2

You have to get rid of h first though.

Answer 3

If the sum of the length of the base side and the height is 9 <-----this already gives you an equation.

Answer 4

Give an equation for that statement?

Answer 5

so it would be \[x + h = 9\] and then you'd solve that for h, \[h = 9 - x\] then put that in for h in the original equation for V?

Answer 6

And then get the first & second derivatives?

Answer 7

Yep. replace h first before doing that.

Answer 8

Then you can differentiate with respect to x.

Answer 9

And you find the S.P's (stationary points) by letting the first derivative equal to zero.

Answer 10

Find the x-values for the stationary points, then sub the x-value into the original equation. Once you do that, sub the x-values into the second derivative to see whether it's a maximum or minimum. If the second derivative is less than zero, it's a maximum if it's larger/greater than zero, it's a minimum.

Answer 11

could you rewrite that equation for the volume. Could you draw it out so I can see it more accurately.

Answer 12

Just to check my math so I don't mess it up, if \[h = 9 - x\] and I'm plugging that into \[V = 1/3x^2h\] it would be \[V = 3x^2-1/3x^3\] right?

Answer 13

Is it
\[\frac{1}{3}x^2h\]
or
\[\frac{1}{3x^2h}\]

Answer 14

\[\frac{ 1 }{ 3 } x^2h\]

Answer 15

Okay.
\[\frac{ dV }{ dx }=\frac{ 18x-3x^2 }{ 3 }\]
\[\frac{ d^2V }{ dx^2 }=-\frac{ 6x }{ 3 }\]

Answer 16

@alaskamath

Answer 17

At S.Ps
dV/dx=0
\[18x-3x^2=0\]
\[6x-x^2=0\]
\[x(6-x)=0\]
\[x=0, 6\]
You can't have a length of zero so of course the length is 6.
\[x=6\]
Sub that into the original equation to find the volume.

Answer 18

And if you sub in any positive x-value into the second derivative it's always negative. So it doesn't matter.

Answer 19

So what was your equation for V after you subbed out h? Is it \[V = \frac{ 1 }{ 3 }x^2(9-x)\]?

Answer 20

@Azteck

Answer 21

Oh sorry about that.

Answer 22

Yeah

Answer 23

sub x=6 into that.

Answer 24

to find the volume.

Answer 25

Okay, thank you!

Answer 26

No worries.