Optimisation question #1

Win_odd Dhamnekar

Junior Member
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Hello Forumites,

1) An isosceles triangle is circumscribed about a circle of radius r. I want to find a height of the triangle that minimises the perimeter of the triangle.
 
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Hi, welcome to the tutoring forum! What have you done thus far? Please share your work and comments. Are you stuck at a particular point, in the exercise? Are there any parts of the exercise you think you don't understand?

Please take time also to check out the forum guidelines. You can start with this summary. Thank you! :cool:
 
A circle with [radius] r is inside the isosceles triangle. Now the perimeter of the isosceles triangle=2*a+b. Now what to do?
Yes, what to do. We have three symbols: a,b,r

We need to express the perimeter in terms of a single variable, so that we can minimize that function using a derivative. One approach, started below, uses an angle measure for this variable. Also, they say, "A circle", so the circle is fixed, whatever its radius. Therefore, our answer will be stated in terms of r.

For example, I can show the smallest possible perimeter is 6∙√3∙r

r ends up being a scaling factor, to adjust for whatever circle we're talking about.

Using symbols a and b won't work because there are many different isosceles triangles that circumscribe a fixed circle, so many different a,b pairs to consider. We need a single variable: x, and I chose to use trigonometry and make x an angle. (There are other approaches.)

I found a function for the triangle perimeter P(x) by:

(1) drawing a picture and dividing the triangle in half (vertically), resulting in two right triangles

(2) using a radius tangent to the isosceles triangle side (i.e., hypotenuse of right triangle) -- this forms a smaller, similar right triangle inside the larger right triangle

(3) defining a shared angle (x) within the two similar right triangles

(4) using symmetry and substitution

Here's a picture.

ISOScirc.JPG

Symmetry: AE=AF and BE=BD=CD=CF

Similar right triangles: ADB (ADB = 90°) and AEO (AEO = 90°)

Note that r=OE=OD=OF

P = 2∙AE + 4∙BD

Now you can start substituting. Use relationships like tan(x)=OE/AE and sin(x)=OE/AO

Your goal is to get:

P(x) = [an expression which contains only symbol r and trigonometric functions of x]

If you get stuck, let us know what's up. Otherwise, please share your result. :cool:

PS: Does your class allow use of technology, for determining derivatives and solving equations?
 
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P= 2*AE+4*BD. We can write 2*AE=2*r cot x. Now how to write 4*BD
Let's see some different ways you're able to express BD in terms of x. That is, what trig functions can you define that involve BD?
 
Let's see some different ways you're able to express BD in terms of x. That is, what trig functions can you define that involve BD?
Hello, So

\(\displaystyle P = 2r \cdot \cot x + \bigg( 4r + 4 \cdot \sqrt{r^2 + r^2 \cdot \cot^2 x} \bigg) \cdot \tan x\)
 
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Yes, your definition for P(x) leads to the same minimum perimeter expression as mine does: 6∙√3∙r

That's encouraging; maybe I didn't goof up.

I had asked earlier whether you're allowed to use technology for finding derivatives and solving equations. You didn't answer.

You need to determine P′(x) next. You can simplify your expression for P(x), by simplifying the radicand and factoring. That would make things easier, if you plan on doing the derivative by hand. :cool:
 
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Hello, So

\(\displaystyle P = 2r \cdot \cot x + \bigg( 4r + 4 \cdot \sqrt{r^2 + r^2 \cdot \cot^2 x} \bigg) \cdot \tan x\)

Then

\(\displaystyle P(x) = 2r \cdot \cot x + \bigg( 4r + 4 \cdot r \cdot cosec(x) \bigg) \cdot \tan x\)

\(\displaystyle P(x) = 2r \cdot \cot x + 4r \cdot \tan x + 4 \cdot r \cdot sec(x) \)

That should not be too difficult to differentiate.

However, finding roots may be different issue....
 
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Thanks, Subhotosh. I missed the trig identity; focused on other simplifications.

The function I had worked out is:

P(x) = 2r [ cot(x) + 2 sec(x) + 2 tan(x) ]

They match now. Looks like we're on the right track!
 
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Use of technology

Thanks, Subhotosh. I missed the trig identity; focused on other simplifications.

The function I had worked out is:

P(r) = 2r [ cot(x) + 2 sec(x) + 2 tan(x) ]

They match now. Looks like we're on the right track!
Hello,

I use the techonolgy in solving maths and stats problems.
 
Optimisation question 1

Excellent. Did you finish? :cool:
Hello,
So i computed 1st derivative \(\displaystyle \frac{4sin^2(x)+2cos^2(x)+4sin(x)}{sin(x)cos(x)}\). Putting P'(r)=0 we get \(\displaystyle (1+sin^2 (x))=-2sin(x)\). Now what to do?
 
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The derivative is not correct. I made a mistake in my notation, mixing up P(r) and P(x).

We're using P(x) to find the minimum perimeter, as a function of angle x.

We need to differentiate with respect to x.

We can assume a unit circle, at this point. (We can write height function h(r) in terms of r, at the end.)

My apologies. :-|
 
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The derivative is not correct. I made a mistake in my notation, mixing up P(r) and P(x).

We're using P(x) to find the minimum perimeter, as a function of angle x.

We need to differentiate with respect to x.

We can assume a unit circle, at this point. (We can write height function h(r) in terms of r, at the end.)

My apologies. :-|
Hello,
After calculating first derivative,I made \(\displaystyle \frac{4rsin^3(x)+4rsin^2(x)-2rcos^2(x)}{\sin^2x\cos^2x}=0\) Now in this equation there are two variables namely r and x.Now what to do?
 
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… first derivative …\(\displaystyle 4rsin^3(x)+4rsin^2(x)-2rcos^2(x)=0\)
The derivative is still not correct. Are you doing it by hand? Please check the function P(x) that you're differentiating. (Subhotosh and I have each posted our versions.)

r is a scaling factor, so it's a constant parameter in our function; it's also going to be in the final answer. Right now, it's just coming along for the ride. As I mentioned, you can set it equal to 1 at this point (unit circle), if you like. :cool:
 
The derivative is still not correct. Are you doing it by hand? Please check the function P(x) that you're differentiating. (Subhotosh and I have each posted our versions.)

r is a scaling factor, so it's a constant parameter in our function; it's also going to be in the final answer. Right now, it's just coming along for the ride. As I mentioned, you can set it equal to 1 at this point (unit circle), if you like. :cool:

Hello, Please read post #13. I have edited it with correct derivative. If it is still incorrect,then show me your derivative.
 
Hello,
After calculating first derivative,I made \(\displaystyle \frac{4rsin^3(x)+4rsin^2(x)-2rcos^2(x)}{\sin^2x\cos^2x}=0\) Now in this equation there are two variables namely r and x.Now what to do?

Your expression for P'(x) is correct. Remember, in this case 'r' is NOT a variable!

You will be trying to optimize the perimeter of the triangle - given a circle of radius 'r'.
 
The root of the above equation is (0≤x≤π/2) x = π/6

So the circumscribing triangle is an equilateral triangle.
 
… Please read post #13. I have edited it with correct derivative.
Very good.

Now, you already understand why a function f(x) has minimum(s) and/or maximum(s) at x-values where the first derivative is 0 (or undefined), yes? Those values of x are "critical values".

Therefore, you need to solve the equation P′(x) = 0, to find the critical values, because we're trying to minimize P(x). :cool:
 
Final answer

Very good.

Now, you already understand why a function f(x) has minimum(s) and/or maximum(s) at x-values where the first derivative is 0 (or undefined), yes? Those values of x are "critical values".

Therefore, you need to solve the equation P′(x) = 0, to find the critical values, because we're trying to minimize P(x). :cool:
Hello,
So the equilateral triangle in which a circle of radius r is inscribed must have height \(\displaystyle \sqrt{a^2-(\frac{a}{2})^2}\) in order to minimise it's perimeter where a is the one of the side of that triangle.
 
… must have height \(\displaystyle \sqrt{a^2-(\frac{a}{2})^2}\) in order to minimise it's perimeter …
It looks like you've skipped ahead.

We don't know side a, so the expression above won't give us the height. We're given a circle with radius r, so that's the only value that we "know". In order to write a height function h(r), we need to express the height in terms of r. Can you do this?

Also, it's not the triangle's height that minimizes its perimeter. You discovered, by finding the critical value \(\displaystyle \pi\)/6, that the triangle's perimeter is minimized when sides a and b are chosen to be equal. In other words, the (circumscribed) isosceles triangle's perimeter is minimized when it's an equilateral triangle.

I would refer back to the diagram. Two line segments comprise the height. We now have expressions for each of them, in terms of r and x. (You found the value of x that minimizes the perimeter; use it). :cool:
 
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