Optimisation: An ornamental container consists of a cuboid-shaped hole inside a sphere.

Look at the cross-section by a vertical plane which contains two diagonally opposite vertical edges of the cuboid.
 
For part (a), refer to the 3D diagram.
Note the right triangle with hypotenuse R, vertical leg y/2, and horizontal leg [imath]\dfrac{\sqrt{2}}{2} \cdot x[/imath]
 
skeeter said:
For part (a), refer to the 3D diagram.
Note the right triangle with hypotenuse R, vertical leg y/2, and horizontal leg [imath]\dfrac{\sqrt{2}}{2} \cdot x[/imath]
Click to expand...
How are you getting that value of the horizontal leg?

I have used Pythagoras and 2r as the hypotenuse, with the vertical leg as y and the horizontal leg as x, however when rearranging this equation I am getting x^2=2r^2-y^2. Where am I getting the (1/2)y^2 from?
 
I think I know what I have been doing wrong and it’s by using the x as the horizontal leg. Should I be using the formula xsqrt(2)?
HATLEY1997 said:
How are you getting that value of the horizontal leg?

I have used Pythagoras and 2r as the hypotenuse, with the vertical leg as y and the horizontal leg as x, however when rearranging this equation I am getting x^2=2r^2-y^2. Where am I getting the (1/2)y^2 from?
Click to expand...
 
HATLEY1997 said:
How are you getting that value of the horizontal leg?

I have used Pythagoras and 2r as the hypotenuse, with the vertical leg as y and the horizontal leg as x, however when rearranging this equation I am getting x^2=2r^2-y^2. Where am I getting the (1/2)y^2 from?
Click to expand...
This may help:

mathcountsnotes.blogspot.com

Face Diagonal and Space Diagonal of a Rectangular Prism

Face diagonal and space diagonal of a cube Ways to calculate face and space diagonal. Each side of the cube is x units long. ...
mathcountsnotes.blogspot.com mathcountsnotes.blogspot.com
 
blamocur said:
Look at the cross-section by a vertical plane which contains two diagonally opposite vertical edges of the cuboid.
Click to expand...
I think part of the confusion may be that this image is not through diagonally opposite edges, but through the front face:
1707496393292.png​
The circle is not a great circle on the sphere, and its diameter is not 2R.

If we imagine this to be a section through the center of the sphere, then the bottom edge would be not [imath]x[/imath], but [imath]x\sqrt{2}[/imath]:

1707496776749.png​
HATLEY1997 said:
I think I know what I have been doing wrong and it’s by using the x as the horizontal leg. Should I be using the formula xsqrt(2)?
Click to expand...
Yes, this would be one way to do it. @skeeter is doing something similar, but working only with R, from the center to a vertex, rather than with the whole cuboid and diagonal/diameter 2R.
 
I understand my first step for part c is finding the derivative, however having both the r and y variables is throwing me off finding the stationary points, any advice?IMG_0375.jpeg
 
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HATLEY1997 said:
I understand my first step for part c is finding the derivative, however having both the r and y variables is throwing me off finding the stationary points, any advice?
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You have V in terms of R and y. But R is a constant, not a variable. It shouldn't get in the way.

1707766957796.png​
 
Thank you!

Am I right in thinking the stationary points are 2r/sqrt(3) and -2r/sqrt(3)?
 
HATLEY1997 said:
Thank you!

Am I right in thinking the stationary points are 2r/sqrt(3) and -2r/sqrt(3)?
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Yes, except that y can't be negative.

And R is uppercase. In math, that is often significant (e.g. R and r might mean outer and inner radii), so you should develop the habit of using variables exactly as given!
 
Dr.Peterson said:
Yes, except that y can't be negative.

And R is uppercase. In math, that is often significant (e.g. R and r might mean outer and inner radii), so you should develop the habit of using variables exactly as given!
Click to expand...
Thank you! To show that this is the maximum, should I use 2nd derivative test?

This would give 2R^2-3y (how do I show this is equal in to a negative value though?)

To find the maximum volume, would this be 2R^2-1/2y^3 * 2R/sqrt(3) ?
 
HATLEY1997 said:
Thank you! To show that this is the maximum, should I use 2nd derivative test?

This would give 2R^2-3y (how do I show this is equal in to a negative value though?)

To find the maximum volume, would this be 2R^2-1/2y^3 * 2R/sqrt(3) ?
Click to expand...
To apply the second derivative test, you just need to find the second derivative (which is not yet correct!), and then put your value for y into the formula for the second derivative. It will be obvious whether it is negative! (It will simplify to a numerical multiple of R.)

To find the max volume, put the value for y into the formula for V. What you show is not correct. (It will simplify to a multiple of R^3, and there will be no y in the answer, since that will have been replaced.)
 
Dr.Peterson said:
To apply the second derivative test, you just need to find the second derivative (which is not yet correct!), and then put your value for y into the formula for the second derivative. It will be obvious whether it is negative! (It will simplify to a numerical multiple of R.)

To find the max volume, put the value for y into the formula for V. What you show is not correct. (It will simplify to a multiple of R^3, and there will be no y in the answer, since that will have been replaced.)
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Is it 4R-3y?
 
Perfect thank you

Am I right in saying this is equal to -6R/sqrt(3) ?

And the maximum volume is 8R^3/3sqrt(3) cm^2 ?
 
HATLEY1997 said:
Am I right in saying this is equal to -6R/sqrt(3) ? [ . . . ]
Click to expand...


Back in your post # 3, with your positive stationary value, but with capital R, \(\displaystyle \ y = \ \dfrac{2R}{\sqrt{3}} \ cm.\)

The second derivative is \(\displaystyle \ -3y, \ \) which you gave in your post # 8, as the derivative with the R-variable is zero.

What is the sign of -3y with this y-value substituted into it? Does this sign indicate that the y-value is a maximum?

You must substitute the y-value in two places in the volume expression, given in part (b), and do some simplifying:

\(\displaystyle V \ = \ \bigg[\dfrac{2R^2}{1}\bigg(\dfrac{2R}{\sqrt{3}}\bigg) \ - \ \dfrac{1}{2}\bigg(\dfrac{2R}{\sqrt{3}}\bigg)^3\bigg] cm.^3\)

\(\displaystyle V \ = \ ? \ cm.^3 \)
 
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lookagain said:
Back in your post # 3, with your positive stationary value, but with capital R, \(\displaystyle \ y = \ \dfrac{2R}{\sqrt{3}} \ cm.\)

The second derivative is \(\displaystyle \ -3y, \ \) which you gave in your post # 8, as the derivative with the R-variable is zero.

What is the sign of -3y with this y-value substituted into it? Does this sign indicate that the y-value is a maximum?

You must substitute the y-value in two places in the volume expression, given in part (b), and do some simplifying:

\(\displaystyle V \ = \ \bigg[\dfrac{2R^2}{1}\bigg(\dfrac{2R}{\sqrt{3}}\bigg) \ - \ \dfrac{1}{2}\bigg(\dfrac{2R}{\sqrt{3}}\bigg)^3\bigg] cm.^3\)

\(\displaystyle V \ = \ ? \ cm.^3 \)
Click to expand...
It is a negative sign, therefore indicating it is a maximum (3*2R/sqrt(3)), which I believe would be equal to -6R/sqrt(3)?

Is the maximum value 8R^3/3sqrt(3) cm^3 ?
 
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