Re: cone inside sphere problem - find the dimensions of the cone of max volume


"jdavca" <msfca@xxxxxxxx> wrote in message
news:26801510.1133296839363.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxxxxx
> Problem:
> The vertex of a right circular cone and the circular edge of its base lie
on the surface of a sphere. The sphere has a radius of 5 feet. Find the
dimensions of the cone of maximum volume that can be fitted into the sphere.
>
> I'm really not sure what to do with this one!
>
> I appreciate anyone who can help me with this...

Think about why this is equivalent to inscribing the largest right triangle
inside the right portion of a circle of radius 5 feet. The equation of the
circle is x^2 + y^2 =25. Actually the right side has the equation x = sqrt
(25 - y^2). Now one vertex of the triangle is at (0,5). Another vertex is at
(0, y) where -5<y<5. The third side is at (x,y) where x^2 + y^2 = 25, ie
this 3rd point lies the circle. Not in the last 2 points those y-values are
equal!
Now the area of the triangle A = .5(base)(height)= .5(x)(5-y). Now you want
to maximize A but it is a function of 2 variable and you only know how to
max a function in one variable. So write A as A = .5(sqrt (25 - y^2))(5-y).
Now using the product rule and the constant times function rule for
derivative you take the derivative of A wrt y and see when this A' = 0 .....
Hope this helps.


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