Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :

Solution

Maths sir · 5 steps

Use the fixed total area to tie r to x, then set the derivative of the total volume to zero.

1. Area and volume. The cuboid's area is 2(8x² + 20x² + 10x²) = 76x². A closed hemisphere has curved area 2π r² plus a flat base π r², so 76x² + 3π r² = k. The total volume is V = 40x³ + 2/3π r³.
2. Link r to x. Differentiate the area condition with respect to x: 152x + 6π rdr/dx = 0, so dr/dx = -76x/(3π r).

   dV/dx = 120x² + 2π r²dr/dx = 120x² - 152/3xr = 0

3. Solve the ratio. Since x ≠ 0, divide by x: 120x = 152/3r, so 360x = 152r. This gives x/r = 152/360 = 19/45.
4. Nature of point. This is the only stationary point. Strictly, dV/dx changes from negative to positive there, so it gives the least total volume. The question means this stationary point.
5. So. The required ratio is x : r = 19 : 45, option (b).

The idea to keep. Fixed total area with two sizes: differentiate the area condition to get dr/dx, put it into dV/dx and set it to zero. A closed hemisphere has area 3π r², not 2π r².

Answer from the NTA final answer key. Solution by PrepWiser.