Calculus Optimization Problem?

Question

What is the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius r? (The circle is inside the triangle). No numbers are given.

yeah i thought it would be equilateral intuitively, but how did you get that?

that is, the problem must be solved without the knowledge that the triangle is equilateral, and solely with the knowledge that it's isosceles.

Ian, your solution works great. I have a question though. I'm having trouble following how you got from

A²/r² = b³r²/(b - 2r)
to the derivative
dA²/db = 2b²r²(b-3r)/(b-2r)²

When I try it myself my answer is slightly different. Could you please explain that further?

Answer 1

Ok I'll start from an isocele's triangle and see where it gets us.
Draw your isoceles triangle around your circle. Draw a line along the line of symmetry through the triangle and circle. It will cut both exactly in half giving a right angle triangle and a semicircle.
Label the point at the angle you just halved A.
Label the right angle of the triangle B.
Label the third corner C.
Let D be the point where the circle touch the line segment AC.
And label the centre of the circle O.

Now OD = OB = r
Let the angle at A be α.
By symmetry BC=CD and let them both equal h (height of the triangle)
Let AB = b (base of the triangle)
Now the area of interest is 2 * 1/2 * b * h = b*h

Using trig:
tanα= h/b ... (in triangle ABC)
sinα = r/(b-r) ... (in triangle ADO)
Now cot²α + 1 = cosec²α ... (trig identity)
So (b/h)² + 1 = ((b-r)/r)²
b²/h² = ((b²-2br+r²)/r)² - 1
b²r²/h² = b² - 2br
br²/h² = (b - 2r)
h²/br² = 1/(b - 2r) ... (equation 1)
b²h²/r² = b³/(b - 2r)
A²/r² = b³/(b - 2r)
A²/r² = b³r²/(b - 2r)
Maximising A² will maximise the area.
r is a constant so A² only depends on b.
dA²/db = 2b²r²(b-3r)/(b-2r)²
This will be equal to zero when either b = 0 or b = 3r.
b = 0 obviously is minimum so ignoring it b =3r.
Sub back into equation 1
h²/br² = 1/(b - 2r)
h²/3r³ = 1/(3r - 2r)
h²/3r³ = 1/r
h² = 3r²
h = √3r
So maximum possible area in terms of r is: b*h = 3√3r².
(This corresponds to the equaliteral case already shown above.)

P.S. There is probably a shorter way you can go about it as I'm doing this at midnight. You should be able to use this to make a simpler solution. Steps:
- Define stuff.
- Find the area in terms of one thing which can vary and r.
- Differentiate.
- Find maximum.

Update: Thought of an some less involved algebra compared to what I originally posted.
Draw it as above and we had:
tanα= h/b ... (in triangle ABC)
sinα = r/(b-r) ... (in triangle ADO)
So
(b-r)sinα = r
b-r = r/sinα
b = r/sinα + r
b = r(1/sinα + 1)
And
h = btanα
h = r(1/sinα + 1)tanα
As A = b*h
A = r²(1/sinα + 1)²tanα
A = r²(cosecα + 1)²tanα
Then take the derivative with respect to α
dA/dα = r² [ 2(cosecα + 1)(-cotαcosecα)tanα + (cosecα + 1)²sec²α ]
dA/dα = r²(cosecα + 1) [ -2cosecα + (cosecα + 1)sec²α ]
Setting equal to zero gives:
cosecα + 1 = 0 or -2cosecα + (cosecα + 1)sec²α = 0
cosecα + 1 = 0 gives α = -π/2 which is obviously a minimum.

-2cosecα + (cosecα + 1)sec²α = 0 ... ( timing by sinαcos²α )
-2cos²α + 1 + sinα = 0
Now cos²α = 1 - sin²α so:
-2(1-sin²α) + 1 + sinα = 0
-2 + 2sin²α + 1 + sinα = 0
2sin²α + sinα - 1 = 0
(2sinα - 1)(sinα + 1) = 0
So α = -π/2 or α = π/6
Again α = -π/2 is a minimum and so α = π/6 is a maximum.
So we can see its a equilateral triangle. Subbing α = π/6 back into A gives:
A = r²(cosecπ/6 + 1)²tanπ/6
A = r²(2 + 1)²/√3
A = 3√3r²
Which is same as before.

Answer 2

Draw a circle with radius r. From the centre of the circle draw three lines to intersect the circle at 120° from each other. Where these lines intersect the circle, will be the tangential points of the triangle. Draw in the three lines of the triangle.
Take any two lines that has been drawn from the centre of the circle, and where they intersect the circurfence of the circle join these two points together with a straight line. Now you have a little triangle within the triangle. The angles are 120°, 30° and 30° (Trig. theory). This implies now, that the outside triangle is equilateral, so it's internal angles are 60°. The length of the side of this triangle will be half the length of the big outside triangle. So calculate this half length.

Working the calculation from the centre of the circle.
Sin Θ = opp./ Hyp.
Sin 60° = x / r
r Sin 60° = x
But 2x is the length of the line of the small triangle.
2x = 2r Sin 60° (This is half the length of the big triangle).

Area of Triangle = ½b*h
Area of Triangle = (2r Sin 60°)(4r Sin 60°).

Answer 3

The triangle is equilateral.

Let
a = length of side of triangle
h = height triangle
A = area of triangle
r = radius of incircle

A = (1/2)ah = (1/2)a [(√3/2)a] = (√3/4)a²

Now find a in terms of r.

Drop a perpendicular from a vertex of the triangle to the other side. It will divide the other side in half with a right angle. Drop a second perpendicular for another vertex. They will intersect in the center of the triangle (which is also the center of the incircle).

Now we have a 30° 60° 90° triangle. The short leg is r because it is the radius of the circle. The long leg is (√3)r. The length of one side of the triangle is twice the long leg.

a = 2*(√3)r = (2√3)r

The area of the triangle is:

A = (√3/4)a² = (√3/4)*[(2√3)r]² = (√3/4)(12r²)
A = (3√3)r²

Answer 4

1) It is simple; ==> First decide the objective of the problem; here it is paper dimensions, so that its area will be minimum; then next comes under what constraints this has to be done? ==> Well, here On the selected paper, you are leaving some margins all over and arriving at the area for printing, which is given some fixed value. 2) Now, we are clear what we want; How to proceed for building mathematical equations, so that it can be solved for getting the end answer: 3) Here we are given the printing area as 50 sq in which is constant; hence let us start from this data; ==> Let the dimension of the printing area be x in (height wise) by y in (width wise) ==> Printing area = xy = 50; == y = 50/x -------(1) 4) There is a margin of 4 in each at top and bottom are provided; hence overall height of the paper is "x + 8" in; Similarly a margin of 2 in is provided on either sides; ==> Overall width = y + 4 in 5) Hence the area of the paper is = (x+8)(y+4) 6) Substituting for y from equation (1), A (x+8)(50/x + 4) = 50 + 400/x + 4x + 32 7) Hence the function to be minimized is: A = 82 + 4x + 400/x 8) Differentiating this, A' = 0 + 4 - 200/x^2 = 4 - 400/x^2 9) Equating A' = 0, x^2 = 100; ==> x = +/- 10 in 10) But a dimension cannot be negative, hence we consider only + 10; So x = 10 in and y = 5 in 11) However we need to verify,whether this is minimum or maximum; for which we will apply 2nd derivative test; So again differentiating, A'' = 800/x^3; at x = 10, A'' is > 0; hence it is minimum Thus we conclude for the printing the area to be 50 sq in, under the given constraints of margins, the outer size of the paper must be 18 in by 9 in; So outer area is = 162 sq in. Wish you are explained; have a nice time.

Answer 5

the triangle is equilateral and isosocles
ok
I got
(r + (r/cos30)) * (r * sin 30/cos 30)

Answer 6

There's a nice solution at http://cs.nyu.edu/~gottlieb/tr/2005-jul-2-2.pdf which can be regarded as a "proof by calculus", if thats what you are looking for! Nice question, thanks!