optimization: cylinder inscribed in a sphere
Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a sphere of radius a
so for the main equation that we will differentiate, i determined that
V(of cylinder) = (pi)(r^2)(h)
and for the connector, i made the picture of the spere and the cylinder flat so that it was two-dimensional and it became a rectangle inscribed in a circle.
i looked at this picture and decided that i had to somehow connect the cylinder to the sphere so i decided to use the equation of a circle as a connector(was this right?)
connector: r^2= x^2 + y^2
from there i found that y=(r^2-x^2)^(1/2) for later use
after this i dont even know where i went....i think i accidentally mixed up some x's and y's that i labeled before and got an illegitimate answer. so basically im confused.
V = pi*r<sup>2</sup>h
x = radius , R = fixed radius of sphere,
h = 2(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>
V = 2pi*x<sup>2</sup>(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>
now ... find dV/dx and find the value of x that maximizes the volume, then find h.
Originally Posted by skeeter
also, isnt the x that we are using for the radius of the cylinder and the x we are using in the equation of the height two different x's?
center a circle of radius R at the origin.
inscribe a rectangle inside the square such that its sides are parallel to the coordinate axes.
rotate the rectangle and circle about the y-axis ... you get a cylinder inscribed in a sphere.
horizontal base of the rectangle has a length = 2x ... that would form the diameter of the cylinder, so the radius of the cylinder = x
vertical distance from the x-axis to the circle along the vertical side of the rectangle is y = (R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>
the total vertical distance, or height of the cylinder, is twice this distance ...
h = 2y = 2(R<sup>2</sup> - x<sup>2</sup>)<sup>1/2</sup>
now ... substitute those expressions, all in terms of x, into the cylinder volume equation.