# find the circular cylinder of greatest volume that can be inscribed in a right circular cone with altitude 20 cm and diameter of base 10 cm.

so at a height h, the radius is 5(1-h/20)

check that.

VolumeCone= PIr^2 H
= PI h * 25(1-h/20)^2

find dV/dh, set to zero, solve for h.

A small furniture company calculates that it can sell 100x tables if the price is set at 200-4x dollars per?
table. the cost is as follows.
C(x) = 6000 + 10000x + 4000/3x.root x . find the optimum of tables and maximum possible profit.
calculus
word to word
due tomrrow
plz

8 months ago

## To find the circular cylinder of greatest volume that can be inscribed in a right circular cone, we can start by visualizing the problem.

We have a right circular cone with an altitude (height) of 20 cm and a diameter of the base of 10 cm. Let's label the radius of the base of the cone as R.

Now, let the height of the cylinder be h and its radius be r. Since the cylinder is inscribed in the cone, the cylinder must have its height and radius equal to the cone at the same level.

Let's consider a cross-section of the cone and the inscribed cylinder. We can see that the radius of the cone at any given height h is smaller than the radius of the base, and the radius of the cylinder should be equal to the radius of the cone at that same height h.

Using similar triangles, we can find the relationship between the radius of the cone and the height. At height h, the radius of the cone can be given by:

r = R * (1 - h/20)

Now, let's find the volume of the cylinder. The volume of a cylinder is given by:

VolumeCylinder = Ï€ * r^2 * h

Substituting the relationship between r (the radius of the cylinder) and R (the radius of the cone), we get:

VolumeCylinder = Ï€ * (R * (1 - h/20))^2 * h

Now, to find the cylinder of the greatest volume, we can differentiate VolumeCylinder with respect to h and set it to zero:

dV/dh = 0

Let's differentiate and solve:

dV/dh = Ï€ * (R^2) * (1 - h/20) * (2 - h/10) = 0

Simplifying the equation, we have:

(1 - h/20) * (2 - h/10) = 0

Expanding and rearranging, we get:

2 - h/20 - 2h/10 + h^2/200 = 0

h^2/200 - h/20 - h/10 + 2 = 0

Now, solving this quadratic equation, we will get two values for h. We need to consider the positive value of h since we are dealing with lengths.

Once we have the value of h, we can substitute it back into the equation for the radius of the cone (r = R * (1 - h/20)) to get the radius of the cylinder at that height.

Finally, we can substitute the values of h and r into the volume equation to get the maximum volume of the inscribed cylinder.

Remember to check your calculations and units to obtain accurate results.

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