#### To determine

**To find:**

Value of r

#### Answer

25

#### Explanation

**1) Concept:**

Use the equation for the volume of cap with height h and subtract the volume of cap of the larger sphere from the volume of hemisphere of the small sphere. Then maximize this function according to the first derivative test.

**2) Calculation:**

Let R=1 be the radius of the sphere.

Consider the portion shaded in gray color to be the cap of the sphere of height h.

Therefore, the volume of the cap of the sphere is

V=13πh2(3r-h)

To calculate h, consider the triangle ∆ABC

1-h2+r2=1

1-h2=1-r2

1-h=1-r2

h=1-1-r2

Vcap=13π1-1-r223-1-1-r2

Vcap=13π1-21-r2+1-r22+1-r2

Vcap=13π2-21-r2-r22+1-r2

Vcap=13π4-41-r2-2r2+21-r2-21-r2-r21-r2

Vcap=13π4-41-r2-2r2+21-r2-2+2r2-r21-r2

Vcap=13π2-21-r2-r21-r2

According to the given condition, maximize the area of region shaded in red.

To find the volume of the shaded region, subtract volume of the cap of the large sphere from the volume of the hemisphere.

Volume of hemisphere=23πr3

V=Vhemisphere-Vcap

V=23πr3-π32-21-r2-r21-r2

V=π32r3-2-21-r2-r21-r2

V=π32r3-2+21-r2+r21-r2

V=π32r3-2+(2+r2)1-r2

Differentiating with respect to r,

dVdr=ddrπ32r3-2+(2+r2)1-r2

dVdr=π36r2+2+r2 -2r21-r2+1-r2 2r

dVdr=π36r2+ (2+r2)(-r)1-r2+2r1-r2

dVdr=π36r2+ -2r-r3+2r(1-r2)1-r2

dVdr=π36r2+ -2r-r3+2r-2r3)1-r2

dVdr=π36r2-3r31-r2

dVdr=π36r21-r2-3r31-r2

Now equate dVdr to zero to get r.

dVdr=0

π36r21-r2-3r31-r2=0

6r21-r2-3r31-r2=0

6r21-r2-3r3=0

21-r2=r

4(1-r2)=r2

4-4r2=r2

4=5r2

r2=45

As r must be positive, taking positive square root,

r=25

Since V'r>0 for 0<r<25 and V'r< 0 for 25<r<1.

V attains a maximum at r=25

Therefore, this is the required value of r at which the required volume is maximum.

**Conclusion:**

r=25