Concept:
To find maxima and minima of a function y = f(x), follow these steps.
Step 1
\(Find\;\frac{{dy}}{{dx}},\;and\;put\frac{{dy}}{{dx}} = 0.\)
Find the value of x and this value is said to be the stationary point, this is a necessary condition to find the extremum value of a function.
Step 2
\(Find\;\frac{{{d^2}y}}{{d{x^2}}}\;\)
Check the value at the stationary point obtained in Step 1.
A function f(x) has a maxima at x = a if f’(a) = 0 and f”(a) < 0
A function f(x) has a minima at x = a if f’(a) = 0 and f”(a) > 0
A function f(x) has no maxima and minima at x = a if f’(a) = 0 and f”(a) = 0.
Calculation:
In the above diagram
θ = semi-vertical angle, h = height, l = slant height, r = radius of the circular base
The volume of the right circular cone is given by -
\(V = \frac{1}{3}\;\pi {r^2}h\)
In ΔABC
r = l × sin θ and h = l × cos θ
\(∴ V = \frac{1}{3}\;\pi \;{\left( {lsin\;\theta } \right)^2}\left( {lcos\;\theta } \right) \Rightarrow \frac{1}{3}\;\pi {l^3}{\sin ^2}\theta \cos \theta \)
Step 1:
\(\frac{{dV}}{{d\theta }} = 0\)
\(\left( {\frac{1}{3}\;\pi {l^3}{{\sin }^2}\theta } \right)\left( { - \sin \theta } \right) + \left( {\frac{1}{3}\;\pi {l^3}\cos \theta } \right)\left( {2\sin \theta \cos \theta } \right) = 0\)
\(\left( {\frac{1}{3}\;\pi {l^3}} \right)\left( { - si{n^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right) = 0\)
\( - \sin \theta \left( {{{\sin }^2}\theta - 2{{\cos }^2}\theta } \right) = 0\)
If sin θ = 0
⇒ θ = 0° which is not possible
∴ sin2 θ – 2cos2 θ = 0
\(\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 2 \ldots \ldots \left( i \right)\)
tan2 θ = 2
\(\tan \theta = \sqrt 2 \)
\(\theta = {\tan ^{ - 1}}\sqrt 2 \)
Step 2:
\(\frac{{{d^2}V}}{{d{\theta ^2}}} = 0\)
\(\frac{{dV}}{{d\theta }} = \left( {\frac{1}{3}\;\pi {l^3}} \right)\left( { - si{n^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right)\)
\(\frac{1}{3}\;\pi {l^3}\left( { - 3{{\sin }^2}\theta \cos \theta + 2\sin \theta \cdot 2\cos \theta \left( { - \sin \theta } \right) + 2{{\cos }^2}\theta \cdot \cos \theta } \right) = 0\)
\(\left( { - 3{{\sin }^2}\theta \cos \theta + 2\sin \theta \cdot 2\cos \theta \left( { - \sin \theta } \right) + 2{{\cos }^2}\theta \cdot \cos \theta } \right) = 0\)
\( - 3{\sin ^2}\theta \cos \theta - 4{\sin ^2}\theta \cos \theta + 2{\cos ^3}\theta = 0\)
\(Using\;\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 2\)
\(( (- 3 )\times (2{\cos ^2}\theta \cos \theta)) - ((4) \times (2{\cos ^2}\theta \cos \theta)) + 2{\cos ^3}\theta = 0\)
\( - 6{\cos ^3}\theta - 8{\cos ^3}\theta + 2{\cos ^3}\theta = 0\)
\( - 12{\cos ^3}\theta \)
Semi-vertical angle θ is less than 90° i.e it lies in 1st quadrant.
cos θ is positive in 1st quadrant
\(∴ \frac{{{d^2}V}}{{d{\theta ^2}}} < 0\)
∴ The volume will be maximum when \({\bf{\theta }} = {\tan ^{ - 1}}\sqrt 2 \)
Open in App
With usual notation, given that total surface area S=πrl+πr2
⇒S=πr√r2+h2+πr2 (∵l=√r2+h2)
⇒Sπr−r=√r2+h2⇒S2π2r2−2Sπ=h2⇒h=√S2π2r2−2Sπ (∵S2π2r2>2Sπ) …(i)
and volume V=13πr2h=13πr2√(S2π2r2−2Sπ)
⇒V=r3√S2−2Sπr2,r2<S2π i.e., 0<r<√S2π
Since, V is maximum, then V2 is maximum
Now, V2=S2r29−2Sπr49,0<r<√S2π
∴ddr(V2)=2rS29−8Sπr39
and ddr2(V2)=2S29−24Sπr29
For maxima put dVdr=0
⇒2rS29−8Sπr39=0⇒r2=S4π⇒d2(V2)dr2<0for r=√S4π
From Eq. (i) h=√S2π2r2−2Sπ=√S2(4π)π2S−2Sπ=√2Sπ
If θ is semi-vertical angle of the cone when the volume is maximum, then in right triangle AOC,
sinθ=r√r2+h2=−√S4π√S4π+2Sπ=1√1+8i.e.,θ=sin−1(13)
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