A cylinder is inscribed in a right circular cone of height 3.5 and radius (at the base) equal to 7. What are the dimensions of such a cylinder which has maximum volume?

Well, the upper radius= lower radius(1-h/3.5)

Volume cylinder=PIr^2*h

volumecylinder= PI (7^2(1-h/3.5)^2 h)
I assume you can maximize that.

I manage to get to that point but maximizing it is the problem.

inscribe rectangle of base r and height h in triangle of height 3.5 and base 7

similar triangles top and bottom
(3.5-h)/r = h/(7-r)
hr = (3.5-h)(7-r) = 24.5 - 3.5 r - 7 h + h r
so
7 h + 3.5 r = 24.5
r = 7 - 3.5 h
(there are easier ways of figuring that out)

now the cylinder volume
v = pi r^2 h
v = pi (49 - 49 h + 12.25 h^2)h
v = pi (49 h -49 h^2 +12.25 h^3)
dv/dh = pi (49 - 98 h + 36.75 h^2)
= 0 for max or min
h = [ 98 +/-sqrt(9604-7203) ] / 98
h = 1 +/- 1/2
h = 1.5 or 0.5
try both of those to see if one is bigger

dv/dh =

7 h + 3.5 r = 24.5

r = 7 - 2 h
(there are easier ways of figuring that out)

now the cylinder volume
v = pi r^2 h
v = pi (49 - 28 h + 4 h^2)h
v = pi (49 h -28 h^2 + 4 h^3)
dv/dh = pi (49 - 56 h + 12 h^2)
= 0 for max or min
h = [56 +/-sqrt(3136-2352) ] / 24
h = 2.33 +/- 1.17
h = 3.5 or .56
3.5 is zero volume so try .56

To find the dimensions of the cylinder that has the maximum volume, we need to use calculus.

Let's denote the radius of the cylinder as r and its height as h. We want to maximize the volume of the cylinder V, given the constraints of being inscribed in the right circular cone.

The volume of a cylinder is given by V = πr^2h. We need to express the height h in terms of the radius r to eliminate one variable.

Considering the geometry of the problem, we can use similar triangles to find the relationship between h and r.

In the given right circular cone, the height of the cone (3.5) is to the radius of the base (7) in the same proportion as the height of the cylinder (h) is to its radius (r). So we have the equation:

3.5 / 7 = h / r

Simplifying this equation, we get:

0.5 = h / r

Now, we can express the volume V only in terms of the radius r:

V = πr^2 * (0.5r)

Simplifying further:

V = 0.5πr^3

To find the maximum volume, we need to take the derivative of V with respect to r, set it equal to zero, and solve for r.

dV/dr = 0.5(3πr^2)

Setting dV/dr equal to zero:

0.5(3πr^2) = 0

3πr^2 = 0

r^2 = 0

Since r^2 cannot be equal to zero, there is no critical point when taking the derivative. However, we need to check the endpoints of the interval to determine if there is a maximum.

The radius r cannot exceed half the base of the cone (7/2 = 3.5) because the cylinder must be inscribed in the cone. Therefore, the only endpoint we need to consider is r = 3.5.

Calculating the volume at the endpoint:

V(3.5) = 0.5π(3.5)^3 = 42.4375π

Now we compare this volume to the value of V at the critical point, which we found to be zero. Since the volume at the endpoint is positive, we conclude that the maximum volume is attained at the endpoint.

Therefore, the dimensions of the cylinder with maximum volume are:

Radius = 3.5
Height = 0.5 times the radius = 1.75