Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is frac2R

Question

\frac{2R}{\sqrt{3}}

CBSECBSE Class XII Board 2019Subjective· 6mImportance★★★★★

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Concept understanding — Lagrange Multipliers

Lagrange Multipliers: Optimising Under a Constraint

To find the highest point on a hill you just walk uphill until every direction goes down — that is unconstrained optimisation. Now suppose you must stay on a footpath through the park. The highest point on the hill may lie off the path, so you want the highest point that lies on the path. That is constrained optimisation, and Lagrange multipliers are the tool for it: maximise or minimise $f(x,y)$ subject to a constraint $g(x,y)=0$ .

The Idea: Two Gradients Line Up

At the best point on the constraint curve, you cannot move along the curve and still increase $f$ — otherwise you would keep going. So the direction of steepest ascent, $\nabla f$ , has no component along the curve; it points straight across it. The curve's own normal direction is $\nabla g$ . Hence at the optimum the two gradients are parallel:

\nabla f = \lambda\, \nabla g

The scalar $\lambda$ is the Lagrange multiplier. It measures how much the optimal $f$ would change if the constraint were loosened by one unit.

Note

This is the geometric picture, not a formal proof — but it is the picture to keep in your head.

The Method

To find the extrema of $f(x,y)$ subject to $g(x,y)=0$ , form the Lagrangian

\mathcal{L}(x,y,\lambda) = f(x,y) - \lambda\, g(x,y)

and solve the system

\frac{\partial \mathcal{L}}{\partial x}=0,\qquad \frac{\partial \mathcal{L}}{\partial y}=0,\qquad \frac{\partial \mathcal{L}}{\partial \lambda}=0.

The third equation just restates $g(x,y)=0$ ; the first two give $\nabla f = \lambda\nabla g$ . Every solution is a candidate — evaluate $f$ at each to decide which is the maximum and which the minimum.

A Worked Example

Find the point on the circle $x^2+y^2=1$ that maximises $f(x,y)=x+y$ .

Here $g = x^2+y^2-1$ , so $\nabla f=(1,1)$ and $\nabla g=(2x,2y)$ . The equations $1=2\lambda x$ and $1=2\lambda y$ give $x=y$ . The constraint then forces $2x^2=1$ , so $x=y=\pm\tfrac{1}{\sqrt2}$ .

At $\left(\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2}\right)$ , $f=\sqrt2$ (the maximum); at $\left(-\tfrac{1}{\sqrt2},-\tfrac{1}{\sqrt2}\right)$ , $f=-\sqrt2$ (the minimum).

Common Mistakes …

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Using Lagrange multipliers, we maximise the cylinder volume $V = 2\pi r^2 h$ subject to the sphere constraint $r^2 + h^2 = R^2$ . The optimal height is $h = \frac{2R}{\sqrt{3}}$ and the maximum volume is $\frac{4\pi R^3}{3\sqrt{3}}$ .

We have a sphere of radius $R$ . Inside it, we inscribe a right circular cylinder — the cylinder's axis passes through the sphere's centre, and its top and bottom faces are parallel circles on the sphere's surface. The cylinder's height $2h$ (total height) and base radius $r$ must satisfy the sphere's equation: the distance from the sphere's centre to any point on the cylinder's rim is $R$ , so $r^2 + h^2 = R^2$ .

The volume of the cylinder is $V = \pi r^2 (2h) = 2\pi r^2 h$ . We want the maximum of $V$ under the constraint $r^2 + h^2 = R^2$ .

Why Lagrange multipliers? Because we have two variables ( $r$ and $h$ ) linked by one equation. Instead of solving for one variable and substituting (which works too), Lagrange multipliers give a symmetric, elegant path — and it's the standard method for constrained optimisation in exams.

Set up the Lagrangian. Define $f(r, h) = 2\pi r^2 h$ (the volume) and $g(r, h) = r^2 + h^2 - R^2 = 0$ (the constraint). The Lagrangian is

\mathcal{L}(r, h, \lambda) = 2\pi r^2 h - \lambda (r^2 + h^2 - R^2).

Take partial derivatives and set to zero.

\frac{\partial \mathcal{L}}{\partial r} = 4\pi r h - 2\lambda r = 0 \quad \Rightarrow \quad 2r(2\pi h - \lambda) = 0.

\frac{\partial \mathcal{L}}{\partial h} = 2\pi r^2 - 2\lambda h = 0 \quad \Rightarrow \quad \pi r^2 - \lambda h = 0.

\frac{\partial \mathcal{L}}{\partial \lambda} = -(r^2 + h^2 - R^2) = 0 \quad \Rightarrow \quad r^2 + h^2 = R^2.

From the first equation, $r = 0$ gives a degenerate cylinder (zero volume), so we take the other factor:

2\pi h - \lambda = 0 \quad \Rightarrow \quad \lambda = 2\pi h.

Substitute $\lambda$ into the second equation.

\pi r^2 - (2\pi h) h = 0 \quad \Rightarrow \quad \pi r^2 = 2\pi h^2 \quad \Rightarrow \quad r^2 = 2h^2.

Use the constraint to find $h$ . $r^2 + h^2 = 2h^2 + h^2 = 3h^2 = R^2 \quad \Rightarrow \quad h^2 = \frac{R^2}{3} \quad \Rightarrow \quad h = \frac{R}{\sqrt{3}}.$ …

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Q.Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R3\frac{2R}{\sqrt{3}}3​2R​. Also find the maximum volume.

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Q.Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2R}{\sqrt{3}}$ . Also find the maximum volume.

Unlock everything free for 14 days