#### To determine

**To find:**

Maximum radius of a ball that causes the greatest volume of water to spill out of the cup

#### Answer

r=hsin θsinθ+cos2θ

#### Explanation

**1) Concept:**

Let S be a solid that lies between y=a and y=b. If the cross sectional area of S in the plane Py through y and perpendicular to the y axis is A(y), where A is a continuous function, then volume of S is

V=∫abAydy

If the cross section formed is the disk, then find the radius of the disk in terms of y and use

Ay=πradius2

**2) Calculation:**

Consider a cone whose tip is at the origin, given height h. Assume that the sphere of radius r is inside the cone at a distance of X from the origin.

Volume of sphere inside the cone can be obtained by revolving the portion circle from X-r to h about x-axis.

sinθ=rX.

The equation of the circle having the center at (X, 0) is

x-X2+y2=r2

So, y=r2-x-X2

By using themethod of washer,

Ax=πr2-x-X22

So, V=∫X-rhπr2-x-X22dx

V=π∫X-rh[r2-x-X2]dx

V=πr2x-x-X33X-rh

V=πr2h-h-X33-r2(X-r)-X-r-X33

V=πr2h-h-rsinθ33-r2rsinθ-r+r33

V=πr2h-h-rsinθ33-r31sinθ-1+r33

V=πr2h-h-rcosec θ33-r3cosec θ-1+r33

V=πr2h-h-rcosec θ33-r3cosec θ-2r33

V=πr2h-h-rcosec θ33-r3cosec θ+2r33

Now, differentiate with respect to r.

dVdr=π2rh+cosecθ·h-r cosecθ2-3r2cosecθ-2r2

dVdr=π2rh+cosecθ·h-r cosecθ2-3r2cosecθ+2r2

dVdr=π2rh+cosecθ·h2-2hrcosec θ+r2cosec2θ-3r2cosecθ+2r2

dVdr=π2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2

Substitute dVdr=0.

π2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2=0

2rh+h2cosec θ-2hrcosec2 θ+r2cosec3θ-3r2cosecθ+2r2=0

Convert the above equation into ax2+bx+c=0 form.

cosec3θ-3cosecθ+2r2+2h-2hcosec2 θr+h2cosec θ=0

Comparing,

a=cosec3θ-3cosecθ+2, b=2h-2hcosec2 θ, c=h2cosec θ

The root of quadratic equation is

x=-b±b2-4ac2a

r=-(2h-2hcosec2 θ)±(2h-2hcosec2 θ)2-4(cosec3θ-3cosecθ+2)h2cosec θ2(cosec3θ-3cosecθ-2)

So,

r=hcosec θ-1 , r=h cosec θ(cosec θ-1)(cosec θ+2)

Substitute the value of r in equation of volume V.

V=πr2h-h-rcosec θ33-r3cosec θ-1-r33

V=πh3cosec θ-12--h33cosec θ-13-h3cosec θ-12-h33cosec θ-13

V=πh3cosec θ-12+h33cosec θ-13-h3cosec θ-12-h33cosec θ-13

V=0

At r=hcosec θ-1, volume V=0

But we need to find maximum value of V, V is a positive function.

Therefore, V reaches its maximum value when r=h cosec θ(cosec θ-1)(cosec θ+2)

Simplifying,

r=h1sinθ1sinθ-11sinθ+2

r=h1sinθ1-sinθsinθ1+2sinθsinθ

r=h1sinθsin2θ1-sinθ1+2sinθ

r=hsin θ1+2sinθ-sinθ-2sin2θ

r=hsin θ1+sinθ-2sin2θ

Since, 1-2sin2θ=cos2θ

r=hsin θsinθ+cos2θ

**Conclusion:**

r=hsin θsinθ+cos2θ