Radius not given?

Calculus Level 4

A Cone is made from a circular sheet by cutting out a sector and gluing the cut edges of the remaining sheet together, as shown in the above figure. θ \theta is the angle of the sector cut out for which the volume of the cone is maximized.

If the value of θ \theta can be written as a ( 1 − b c ) π a\left(1-\frac { \sqrt { b } }{ \sqrt { c } } \right)\pi , where a , b , a,b, and c c are positive integers and b b and c c are not divisible by the square of any prime, determine the value of a + b + c a+b+c .


The answer is 7.

Ahmed Arup Shihab by Ahmed Arup Shihab , 28, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Once the sector is cut from the from the sheet, the remaining circular arc will serve as the circumference of the base of the cone. Letting the radius of the circular sheet be r , r, the remaining arc has a length of 2 π ∗ r − θ ∗ r = r ( 2 π − θ ) . 2\pi*r - \theta*r = r(2\pi - \theta). Letting the radius of the base of the cone be R , R, we then have that

r ( 2 π − θ ) = 2 π ∗ R ⟹ R = r ( 1 − θ 2 π ) = r x r(2\pi - \theta) = 2\pi*R \Longrightarrow R = r\left(1 - \dfrac{\theta}{2\pi}\right) = rx , where x = 1 − θ 2 π . x = 1 - \dfrac{\theta}{2\pi}.

Now the height h h of the resulting cone will be the vertical leg of a right triangle with base length R R and hypotenuse, (the slant height of the cone), of length r . r. Thus h = r 2 − R 2 = r 1 − ( R r ) 2 . h = \sqrt{r^{2} - R^{2}} = r\sqrt{1 - \left(\dfrac{R}{r}\right)^{2}}. The volume of the cone is then

V = 1 3 ∗ π R 2 ∗ h = π r 3 3 x 2 1 − x 2 . V = \dfrac{1}{3}*\pi R^{2}*h = \dfrac{\pi r^{3}}{3}x^{2}\sqrt{1 - x^{2}}.

Then by the chain and product rules we find that

d V d θ = d V d x ∗ d x d θ = π r 3 3 ( 2 x 1 − x 2 − x 3 1 − x 2 ) ∗ − 1 2 π , \dfrac{dV}{d\theta} = \dfrac{dV}{dx}*\dfrac{dx}{d\theta} = \dfrac{\pi r^{3}}{3}\left(2x\sqrt{1 - x^{2}} - \dfrac{x^{3}}{\sqrt{1 - x^{2}}}\right) * \dfrac{-1}{2\pi},

which equals 0 0 when 2 x 1 − x 2 = x 3 1 − x 2 ⟹ 2 x ( 1 − x 2 ) = x 3 ⟹ x ( 3 x 2 − 2 ) = 0. 2x\sqrt{1 - x^{2}} = \dfrac{x^{3}}{\sqrt{1 - x^{2}}} \Longrightarrow 2x(1 - x^{2}) = x^{3} \Longrightarrow x(3x^{2} - 2) = 0.

So the critical points occur at x = 0 x = 0 and x 2 = 2 3 ⟹ x = ± 2 3 . x^{2} = \dfrac{2}{3} \Longrightarrow x = \pm \dfrac{\sqrt{2}}{\sqrt{3}}.

Now V = 0 V = 0 at x = 0 x = 0 , and V > 0 V \gt 0 for the other two critical points, so the maximum must occur at one of them. But as we are looking for x > 0 x \gt 0 we conclude that V V is maximized when

x = 1 − θ 2 π = 2 3 ⟹ θ = 2 ( 1 − 2 3 ) π , x = 1 - \dfrac{\theta}{2\pi} = \dfrac{\sqrt{2}}{\sqrt{3}} \Longrightarrow \theta = 2\left(1 - \dfrac{\sqrt{2}}{\sqrt{3}}\right)\pi, and so a + b + c = 2 + 2 + 3 = 7 . a + b + c = 2 + 2 + 3 = \boxed{7}.

6 Helpful 0 Interesting 2 Brilliant 0 Confused

Differentiate V V = 1 3 π \frac13 \pi [ 1 − θ 2 π ] 2 [1 - \frac{\theta}{2 \pi}]^2 1 − ( 1 − θ 2 π ) 2 \sqrt{1 - (1 - \frac{\theta}{2 \pi})^2} directly is all right with

p p = [ 1 − θ 2 π ] 2 [1 - \frac{\theta}{2 \pi}]^2 1 − ( 1 − θ 2 π ) 2 \sqrt{1 - (1 - \frac{\theta}{2 \pi})^2} . I usually don't check whether it is a maximum or a minimum when the logic tells that there ought to be a maximum more obviously. Since θ ≠ 2 π , \theta \neq 2 \pi, we can obtain θ = 2 ( 1 − 2 3 ) π . \theta = 2 (1 - \sqrt{\frac23}) \pi.

2 + 2 + 3 = 7

Answer = 7 \boxed{7}

Lu Chee Ket - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...