Question

Among all right circular cones with a slant height of 27​, what are the dimensions​ (radius and​ height) that maximize the volume of the​ cone? The slant height of a cone is the distance from the outer edge of the base to the vertex.

Answer 1

Answer:

Radius = 22.04 unit,

Height = 15.58 unit

Step-by-step explanation:

Since, the volume of cone,

$V=\frac{1}{3}\pi x^2 h$

Where,

x = radius,

h = height,

Since, the slant height of a cone is,

$l=\sqrt{x^2 + h^2}$

$l^2 = x^2 + h^2$

$\implies h=\sqrt{l^2 - x^2}$

Here, l = 27.

$\implies h = \sqrt{27^2 - x^2}=\sqrt{729 - x^2}----(1)$

So, the volume would be,

$V=\frac{1}{3}\pi x^2 (\sqrt{729 - x^2})$

Differentiating with respect to x,

$\frac{dV}{dt}=\frac{1}{3}\pi (x^2 \times \frac{1}{2\sqrt{729 - x^2}}\times -2x + \sqrt{729 - x^2}\times 2x)$

$=\frac{1}{3}\pi (\frac{-x^3+2x(729-x^2)}{\sqrt{729 - x^2}}$

$=\frac{1}{3}\pi (\frac{-x^3+1458x-2x^3}{\sqrt{729 - x^2}})$

$=\frac{1}{3}\pi (\frac{-3x^3+1458x}{\sqrt{729 - x^2}})$

$=\pi (\frac{-x^3+486x}{\sqrt{729 - x^2}})$

For maxima or minima,

$\frac{dV}{dt}=0$

$\implies -x^3 + 486x = 0$

$x^3 = 486x$

$x^2 = 486$

$\implies x=\pm \sqrt{486}=\pm 22.04$

But side can not be negative,

So, x = 22.04,

Since,

$\frac{dV}{dx}|_{x=20}=297.91\text{Positive}$

While,

$\frac{dV}{dx}|_{x=23}=-219.70\text{Negative}$

Thus, by the first derivative test,

V(x) is maximum at x = 22.04,

Also, from equation (1),

h = 15.58

Hence, for maximising the volume, the radius and height of the cone would be 22.04 unit and 15.58 unit respectively.