Volume of hemisphere
= Half of a sphere's volume
= ½ (4/3 πr³)
= ⅔ πr³


Now, it is given that volume of lid is 1200cm³, and r is a fixed value.

So 1200 = ⅔πr³
r³ = 1200 / ⅔π
r³ = 1800/π
r = ³√(1800/π) = (1800/π)¹/³


Now the radius of the sphere is measured from the centre to any point on its surface.

So the length from the centre (where you see the arrows for x and r meet) to the surface of the lid (top end of the dotted line labelled h) is also r cm.

Now you can use Pythagoras' Theoem since we have a right-angled triangle.


h² + x² = r²
h² + x² = ( (1800/π)¹/³ )²
h² + x² = (1800/π)²/³

x² = (1800/π)²/³ - h²



Volume of cake
= Volume of cylinder of radius x and height h

= πx²h
= π[(1800/π)²/³ - h²]h

= πh[(1800/π)²/³ - h²]

(Shown)

Ahh I didn’t think of using r as pythagoras theorem

Thanks!!

For the next part of the qn, is it possible to simply
h = √[(1/3)(1800/π)^2/3]

V = πh[(1800/π)²/³ - h²]

V = (1800/π)²/³ πh - πh³

dV/dh = (1800/π)²/³ π - 3πh²


When dV/dh = 0,

(1800/π)²/³ π - 3πh² = 0

3πh² = (1800/π)²/³ π

h² = ⅓(1800/π)²/³ = (1800/π)²/³ / 3


h = √ [ (1800/π)²/³ / 3 ]

= ((1800/π)²/³)¹/² / √3

= (1800/π)¹/³ / √3

= (2³ x 5² x 3²)¹/³ / (π¹/³ 3¹/²)

= 2 · 5²/³ · 3²/³ -¹/² / π¹/³

= 2 · 5²/³ · 3^(1/6) / π¹/³

But it should be okay to leave it as

³√(1800/π) / √3

Many thanks!