Maximisation and minimisation in geometry

Optimisation problems in geometry: shapes inscribed under curves and inside solids, using similar triangles or Pythagoras to link the variables, and minimising distance by minimising its square.

A rectangle under a curve

Example 1●●

A rectangle has its base on the $x$-axis and its top two corners on the parabola $y = 12 - x^2$. Find its largest area.

Let the top right corner be $(x, 12 - x^2)$ with $0 < x < \sqrt{12}$. By symmetry the width is $2x$.

$A = 2x(12 - x^2) = 24x - 2x^3$, so $A' = 24 - 6x^2 = 0$ at $x = 2$.

$A'' = -12x < 0$, so this is a maximum: $A = 4 \times 8 = 32$.

The rectangle under y = 12 − x²

The blue line is the top of the rectangle. Slide x and watch the area.

Try this

Check that the area is largest at $x = 2$.

A cylinder inside a cone

Example 2●●●

A cylinder stands inside a cone of base radius $6$ and height $12$, with its base on the cone's base. Find the largest possible volume of the cylinder.

If the cylinder has radius $r$ and height $h$, then $\dfrac{12 - h}{r} = \dfrac{12}{6}$, so $h = 12 - 2r$, with $0 < r < 6$.

Similar triangles in the cross-section: the cone narrows by 1 unit of radius for every 2 units of height.

$V = \pi r^2 (12 - 2r) = \pi(12r^2 - 2r^3)$, so $V' = \pi(24r - 6r^2) = 6\pi r(4 - r)$.

$V' = 0$ at $r = 4$, and $V'' = \pi(24 - 12r) < 0$ there. So the maximum is $V = \pi \times 16 \times 4 = 64\pi$.

The closest point

Example 3●●●

Find the shortest distance from $(0, 2)$ to the parabola $y = x^2$.

For a point $(x, x^2)$ on the parabola, $D^2 = x^2 + (x^2 - 2)^2 = x^4 - 3x^2 + 4$.

$\dfrac{d(D^2)}{dx} = 4x^3 - 6x = 2x(2x^2 - 3)$, which is zero at $x = 0$ or $x^2 = \tfrac32$.

D is smallest exactly when D² is smallest, and D² has no square root.

At $x^2 = \tfrac32$: $D^2 = \tfrac94 - \tfrac92 + 4 = \tfrac74$. At $x = 0$: $D^2 = 4$, which is larger.

The shortest distance is $\sqrt{\tfrac74} = \dfrac{\sqrt7}{2}$.

Flashcards

Shape inscribed under $y = f(x)$

Put a corner at $(x, f(x))$.

Cylinder in a cone

Use similar triangles to write $h$ in terms of $r$.

Shortest distance

Minimise $D^2$.

Practice

Find the largest area of a rectangle under $y = 12 - x^2$ with its base on the $x$-axis.
In that problem, what value of $x$ gives the maximum?
A cylinder inside a cone of radius $6$ and height $12$ has the largest volume when $r = $
In that problem, the cylinder's height is linked to its radius by:
What is the height of that largest cylinder?
Find the smallest value of $D^2$ from $(0, 2)$ to $y = x^2$.
Why minimise $D^2$ instead of $D$?
A rectangle has a diagonal of length $10$. Find its largest possible area.
In Example 1, what is the domain of $x$?