Hello halle,
Let's work this problem in general terms so that we then derive a formula into which we can plug the given data.
First, let's draw a diagram of a cross-section through the axes of symmetry for the cone and cylinder. We will orient our coordinate axes such that the origin as at the center of the circular base of the cone. The radius and height of the cone are $R$ and $H$ respectively, while the radius and height of the cylinder are $r$ and $h$ respectively. Since these parameters and variables represent linear measures, we assume they are non-negative. Both measures of the cone must be positive.
View attachment 1629
Now, our objective function, that function that we wish to optimize, is the volume $V$ of the cylinder:
$$V(h,r)=\pi hr^2$$
As we can see, the point $(h,r)$ is constrained to lie along the line:
$$y=-\frac{H}{R}x+H$$
and thus, our constraint may be given as:
$$g(h,r)=\frac{H}{R}r-H+h=0$$
Let's first use a single variable approach. We may solve the constraint for $h$:
$$h=H-\frac{H}{R}r=\frac{H(R-r)}{R}$$
Substituting for $h$ into the objective function, we obtain the volume of the cylinder in terms of its radius only (recall the radius and height of the cone are parameters, not variables, and so we treat them as constants):
$$V(r)=\pi\left(\frac{H(R-r)}{R} \right)r^2=\frac{\pi H}{R}\left(Rr^2-r^3 \right)$$
Next, we want to differentiate this with respect to $r$ and equate the result to zero to find the critical value(s).
$$V'(r)=\frac{\pi H}{R}\left(2Rr-3r^2 \right)=\frac{\pi H}{R}r(2R-3r)=0$$
Hence, we find two critical values:
$$r=0,\,\frac{2}{3}R$$
To determine the nature of the extrema associated with these critical values, we may use the second derivative test. Computing the second derivative of the volume function with respect to $r$, we find:
$$V''(r)=\frac{\pi H}{R}\left(2R-6r \right)=\frac{2\pi H}{R}\left(R-3r \right)$$
Thus, we find:
$$V''(0)=2\pi H>0$$
Thus the critical value $r=0$ is at a minimum.
$$V''\left(\frac{2}{3}R \right)=\frac{2\pi H}{R}\left(R-3\left(\frac{2}{3}R \right) \right)=-2\pi H<0$$
Thus the critical value $r=\frac{2}{3}R$ is at a maximum. This is the critical value in which we are interested. Thus the dimensions of the cone which maximize its volume are:
$$r=\frac{2}{3}R$$
$$h=\frac{H\left(R-\frac{2}{3}R \right))}{R}=\frac{1}{3}H$$
We may also use a multi-variable technique: Lagrange multipliers. Using this method, we obtain the system:
[math]2\pi rh=\frac{H}{R}\lambda[/math]
[math]\pi r^2=\lambda[/math]
This implies:
[math]2\pi rh\frac{R}{H}=\pi r^2[/math]
Observing that $r=0$ is the critical point associated with the minimal volume, we may divide through by $\pi r$ to obtain:
[math]2h\frac{R}{H}=r[/math]
Now, substituting for $r$ into the constraint, we find:
[math]h+\frac{H}{R}\cdot2h\frac{R}{H}-H=0[/math]
[math]h=\frac{H}{3}\implies r=\frac{2}{3}R[/math]
Hence, from both methods, we find the maximum volume of the cylinder is:
[math]V_{\max}=V\left(\frac{2}{3}R,\frac{H}{3} \right)=\pi\left(\frac{2}{3}R \right)^2\left(\frac{H}{3} \right)=\frac{4}{27}\pi R^2H[/math]
This is $\dfrac{4}{9}$ the volume of the cone.
Now we may plug in our given data:
$$H=6,\,R=\frac{9}{2}$$
to obtain for this problem:
$$r=\frac{2}{3}\cdot\frac{9}{2}=3$$
$$h=\frac{1}{3}\cdot6=2$$
Thus, we have found that the dimensions of the cylinder inscribed within the given cone having maximum volume are a radius of 3 units and a height of 2 units.
