Question

Among all right circular cones with a slant height of 24​, 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. Let V be the volume of the cone. What is the objective function in terms of the height of the​ cone, h?

Answer 1

Answer:

5571.99

Step-by-step explanation:

We need to use the Pythagorean theorem to solve the problem.

The theorem indicates that,

$r^2+h^2=24^2 \\r^2+h^2=576\\r^2=576-h^2$

Once this is defined, we proceed to define the volume of a cone,

$v=\frac{1}{3}\pi r^2 h$

Substituting,

$v=\frac{1}{3} \pi (576-h^2)h\\v=\frac{1}{3} \pi (576h-h^3)$

We need to find the maximum height, so we proceed to calculate h, by means of its derivative and equalizing 0,

$\frac{dv}{dh} = \frac{1}{3} \pi (576-3h^2)$

$\frac{dv}{dh} = 0$ then $\rightarrow \frac{1}{3}\pi(576-3h^2)=0$

$h_1=-8\sqrt{3}\\h_2=8\sqrt{3}$

We select the positiv value.

We have then,

$r^2 = 576-(8\sqrt3)^2 = 384\\r=\sqrt{384}$

We can now calculate the maximum volume,

$V_{max}= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\sqrt{384})^2 (8\sqrt{3}) = 5571.99$