Welcome to our community

Be a part of something great, join today!

Trig problem: max square cut from circle

Bmanmcfly

Member
Mar 10, 2013
42
[SOLVED]Trig problem: max square cut from circle

image.jpg
Since this one is homework related, ill ask this as a procession so that I can make sure that I am grasping the logic.

the question is that with a round log with a diameter of 16cm, what is the strongest log that can be cut where the strength = width * depth^2.

So, I figured the biggest square would use the diameter. I have to use trig functions, so, I have the d as \(\displaystyle d= \sqrt{(16\sin(x))}\). Then I put \(\displaystyle w=16\cos(x)\)

because d is squared relative to w.

does the logic of this make sense so far? (Like the pic)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Have you studied Lagrange multipliers?
 

Bmanmcfly

Member
Mar 10, 2013
42

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, I just wanted to be sure before suggesting that method, as it would be simpler computationally, however, it can still be done. I would approach this problem not in a trigonometric manner, but rather using a coordinate geometry approach.

I would orient my coordinate axes such that the center of the circular cross section of the log is at the origin, and so the perimeter of the cross section is a circle with radius 8 (let all linear measures be in cm) centered at the origin. How can we express the perimeter?

Next, I would let the upper right vertex of the rectangular beam to be cut from the log, be at the point $(x,y)$ in the first quadrant.

Can you now express the width and height of the rectangular cross section, and using the perimeter of the log, then express the strength function in one variable?
 

Bmanmcfly

Member
Mar 10, 2013
42
Okay, I just wanted to be sure before suggesting that method, as it would be simpler computationally, however, it can still be done. I would approach this problem not in a trigonometric manner, but rather using a coordinate geometry approach.

I would orient my coordinate axes such that the center of the circular cross section of the log is at the origin, and so the perimeter of the cross section is a circle with radius 8 (let all linear measures be in cm) centered at the origin. How can we express the perimeter?

Next, I would let the upper right vertex of the rectangular beam to be cut from the log, be at the point $(x,y)$ in the first quadrant.

Can you now express the width and height of the rectangular cross section, and using the perimeter of the log, then express the strength function in one variable?
While I see how you are approaching this, I must use a trigonometric functions to find the dimensions.

Id be much happier not using trig functions myself. Especially that you don't have an angle and only 1 side is known.

So, what I have ATM is \(\displaystyle s=16\cos(x)•(16\sin(x))^2\)

From there, I figure I need to find a max value, so I'll find the derivative of s, \(\displaystyle \frac{ds}{dx}=-\sin^3(x)+8192\cos^2(x)\sin(x)\)

How does this sound sofar
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Your strength function looks good, but your differentiation looks incorrect. I would write:

\(\displaystyle s(x)=4096\sin^2(x)\cos(x)\)

Now, just carry the constant, and differentiate the product of the two trig functions. what do you find?
 

Bmanmcfly

Member
Mar 10, 2013
42
Your strength function looks good, but your differentiation looks incorrect. I would write:

\(\displaystyle s(x)=4096\sin^2(x)\cos(x)\)

Now, just carry the constant, and differentiate the product of the two trig functions. what do you find?
Thanks, would it still be equivalent to simplify 2cosx•sinx giving
\(\displaystyle s(x)=2048\sin(2x)\sin(x)\)
??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you could do that, as you have correctly applied the double-angle identity for sine. (Yes)
 

Bmanmcfly

Member
Mar 10, 2013
42
Yes, you could do that, as you have correctly applied the double-angle identity for sine. (Yes)
Ok, so, I wound up with
\(\displaystyle \frac{ds}{dx}=4096\cos(2x)\sin(x)+2048\sin(2x)\cos(x)\)

Which I plugged into the graphing calc to get a 0at 0.95532.

So, I suppose I should ask here if I could be directed to a simple method to sketch a complex trig equation like this? Better that I remember a method to do this when it comes to test time...

Edit:Ok, sweet, I was handling this correctly, and when I used x at the first 0, it gave me the correct answers that I needed.

The question does remain, is it really just a matter of remembering the y range ?
Thanks for the help...
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write:

\(\displaystyle s'(x)=2048(\sin(2x)\cos(x)+2\cos(2x)\sin(x))=4096 \sin(x)(\cos^2(x)+2\cos^2(x)-1)=\)

\(\displaystyle 4096\sin(x)(3\cos^2(x)-1)=0\)

We know we want \(\displaystyle 0<x<\frac{\pi}{2}\) hence the critical values comes from:

\(\displaystyle x=\cos^{-1}\left(\frac{1}{\sqrt{3}} \right)\approx0.9553166181245092\)

How can we demonstrate using the calculus, that this critical value is at a maximum for the strength function?
 

Bmanmcfly

Member
Mar 10, 2013
42
I would write:

\(\displaystyle s'(x)=2048(\sin(2x)\cos(x)+2\cos(2x)\sin(x))=
4096 \sin(x)(\cos^2(x)+2\cos^2(x)-1)=\)

\(\displaystyle 4096\sin(x)(3\cos^2(x)-1)=0\)

We know we want \(\displaystyle 0<x<\frac{\pi}{2}\) hence the critical values comes from:

\(\displaystyle x=\cos^{-1}\left(\frac{1}{\sqrt{3}} \right)\approx0.9553166181245092\)

How can we demonstrate using the calculus, that this critical value is at a maximum for the strength function?
\(\displaystyle 4096\sin(x)(3\cos^2(x)-1)=0\)

Is it wrong to just divide the \(\displaystyle 4096\sin(x)\) out, since it would be dividing under a 0? It seems valid, but feels too easy.

To your question, in this case the assumption could be made that the first critical point was a max, but otherwise running the second derivative would show if the curve was up or down. Going down would prove it was a max.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you may divide through by that factor since it yields no roots for $x$ in the desired interval.

Also, you are right; the second derivative test is a valid means of determining the nature of the extremum. If the second derivative of the strength function is negative at the critical value, then you know it is concave down there, and so the extremum must then be a maximum.