A cone-roof model with least surface area

Cone-roof surface-area modelGiven cone roof of radius15x and height8x joined to cylinder of radius15x and heightw. Slant height and exposed-surface conclusions are not supplied.8x15xw
Question diagram 1
Cone-roof surface-area modelGiven cone roof of radius15x and height8x joined to cylinder of radius15x and heightw. Slant height and exposed-surface conclusions are not supplied.8x15xw

Given cone roof and cylindrical walls, dimensions8x,15x,w.

[It is given that a right circular cone with base radius rr and height hh has volume 13πr2h\frac13\pi r^2h and curved surface area πrℓ\pi r\ell, where ℓ\ell is the slanted height of the cone.]

The model of a house is made up of three parts.

  • The roof is modelled by the curved surface of a right circular cone with base radius 15x15x cm and height 8x8x cm.
  • The walls are modelled by the curved surface of a cylinder of radius 15x15x cm and height ww cm.
  • The floor is modelled by a circular disc of radius 15x15x cm.

The three parts are joined together as shown in the diagram. The model is made of material of negligible thickness. It is given that the volume of the model is a fixed value k cm3k\,\mathrm{cm}^3, and the external surface area is a minimum. Use differentiation to find the exact values of xx and ww in terms of kk. Simplify your answers. marks

[2022/NYJC/Prelim/I/Q2]

Solution

Optimal dimensions

Optimisation with a fixed volume

A house model has a conical roof (radius 15x15x, height 8x8x), a cylindrical wall (radius 15x15x, height ww) and a circular floor. Its volume is a fixed k cm3k\ \mathrm{cm}^3 and its external surface area is a minimum. Find xx and ww in terms of kk.

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Hint

  1. Volume: cone + cylinder =k=k.

  2. Surface area: cone's curved surface + cylinder's curved surface + floor. The cone's slant height is (8x)2+(15x)2=17x\sqrt{(8x)^2+(15x)^2}=17x.

  3. Use the volume to eliminate ww, differentiate, and solve.

Answer · Optimal dimensions

x=110k/(6π)3x=\frac1{10}\sqrt[3]{k/(6\pi)}, w=125k/(6π)3w=\frac{12}{5}\sqrt[3]{k/(6\pi)}.

k>0k>0; includes the circular floor but excludes the joined cone base.

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Surface area against x (shown for k = 6000π)

The plotted curve is A/π = 800/x + 400x². With this k the minimum is at x = 1.

Position along the curvex = 0.6
Try an example
Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(x)
1.48e+3
Gradient f′(x)
-1.74e+3
negative

Left of x=1 the gradient is negative: increasing x reduces the area.

Selected values are within the original drawing windows.

  • A/π against x
  • Tangent
  • Gradient, and where A increases
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

A=2k15x+400πx2A=\frac{2k}{15x}+400\pi x^2 is large for very small xx (tall thin house) and for large xx (flat wide house): a minimum lies between.

Ideas to try
Where is the tangent horizontal? Is it a minimum or a maximum?