AA HL Optimisation

Worked example 1: Minimise the travel time

Question. A student starts at point \(P\), which is \(300\) m from a straight path. The destination \(Q\) is \(900\) m along the path from the nearest point \(A\). The student walks across grass at \(1.2\text{ m s}^{-1}\), then runs along the path at \(2\text{ m s}^{-1}\).

The student walks from \(P\) to a point \(X\) on the path, then runs from \(X\) to \(Q\). Let \(\theta\) be the angle between \(PX\) and the perpendicular route \(PA\). Find the value of \(\theta\) that minimises the total time.

Solution

Step 1: Identify the target and constraint

The target quantity is the total travel time. The constraints come from the right-angled triangle and from the fixed distance \(AQ=900\text{ m}\).

Step 2: Write the model

Let \(\theta\) be the angle between \(PX\) and the perpendicular route \(PA\). Time is distance divided by speed, so

From the triangle,

and therefore

Step 3: Use the constraint

This is direct substitution (Route A): the trigonometric constraints already express every distance in terms of \(\theta\). Substituting gives

Because \(X\) lies between \(A\) and \(Q\), the contextual domain is

Step 4: Optimise and check

Differentiate:

Set \(T'(\theta)=0\):

Hence

Since \(\sec^2\theta>0\), the sign of \(T'(\theta)\) changes from negative to positive as \(\sin\theta\) passes through \(\frac35\). Therefore the travel time decreases and then increases, so this critical value gives the minimum.

Answer: The minimum time occurs when \(\theta=\arcsin\left(\frac35\right)\).

Worked example 2: A barrier changes the best route

Question. A student starts at point \(P\), which is \(300\) m from the nearest point \(A\) on a straight path. The destination \(Q\) is \(900\) m along the path from \(A\). The student walks across grass at \(1.2\text{ m s}^{-1}\), then runs along the path at \(2\text{ m s}^{-1}\).

The student walks from \(P\) to a point \(X\) on the path and then runs from \(X\) to \(Q\). A barrier requires \(AX\leq120\) m. Let \(\theta\) be the angle between \(PX\) and the perpendicular route \(PA\). Find where the student should reach the path to minimise the total travel time.

Solution

Step 1: Identify the target and constraint

The target quantity is the total travel time. The constraints come from the right-angled triangle, the fixed distance \(AQ=900\text{ m}\), and the barrier restriction \(AX\leq120\text{ m}\).

Step 2: Write the model

From the triangle,

so

Therefore

Step 3: Use the constraint

This is Route A. Substitution gives

The barrier restricts the domain:

Step 4: Optimise and check

Differentiate:

The unrestricted stationary value would satisfy

for which \(\tan\theta=\frac34\). This would give \(AX=300\left(\frac34\right)=225\text{ m}\), so it is not allowed.

On the allowed interval, \(\theta\leq\arctan\left(\frac25\right)<\arcsin\left(\frac35\right)\), and hence \(T'(\theta)<0\). Therefore \(T\) is decreasing throughout the allowed interval, so the minimum occurs at its right-hand endpoint:

Answer: The student should reach the path at the furthest allowed point, \(120\) m from \(A\).

Worked example 3: Maximise a rectangle in an ellipse

Question. A rectangle is centred at the origin with its sides parallel to the coordinate axes. Its vertices lie on the ellipse

Find the maximum possible area of the rectangle.

Solution

Step 1: Identify the target and constraint

The target quantity is the area of the rectangle. The constraint is that every vertex lies on the ellipse.

Step 2: Write the model

Work in the first quadrant, so \(x\geq0\) and \(y\geq0\). The rectangle has width \(2x\) and height \(2y\), hence

The separate constraint is

Step 3: Use the constraint

Solving explicitly for \(y\) would give \(y=4\sqrt{1-\frac{x^2}{36}}\), which introduces an unnecessary square root. We therefore use implicit differentiation (Route B). Regard \(y=y(x)\), so that along the ellipse

Differentiate the constraint with respect to \(x\):

so

The relevant domain is \(0\leq x\leq6\), with \(y\geq0\).

Why this works. The area still contains \(x\) and \(y\), but \(y=y(x)\) along the ellipse. Substituting for \(\frac{dy}{dx}\) now lets us differentiate the area without solving for \(y\).

Step 4: Optimise and check

Differentiate the target quantity and substitute the result from Step 3:

At an interior critical point,

This gives a relationship between the variables, not yet their actual values. We now return to the original constraint from Step 2. Since \(y^2=\frac{4x^2}{9}\), substitute into the ellipse:

Thus

Therefore

At the endpoints of the first-quadrant arc, \((0,4)\) and \((6,0)\), the rectangle has area \(0\). Hence the interior critical point gives the absolute maximum.

Answer: The maximum possible area is \(48\) square units.

Worked example 4: Find the closest and furthest points

Question. Points \((x,y)\) lie on the curve

Find the points on the curve that are closest to the origin and furthest from the origin.

Solution

Step 1: Identify the target and constraint

The target quantity is the distance from the origin. To avoid a square root, we optimise the square of the distance; this gives the same closest and furthest points. The constraint is the given curve.

Step 2: Write the model

Step 3: Use the constraint

Solving the constraint explicitly for \(y\) would introduce quadratic-formula expressions and two branches. We therefore use Route B: regard \(y=y(x)\) wherever the curve is not vertical and differentiate the constraint:

Hence

Along the constraint, the target is the one-variable function

In Step 4, this expression for \(\frac{dy}{dx}\) will be substituted into \(\frac{d}{dx}(D^2)\), eliminating \(\frac{dy}{dx}\) from the derivative of the target quantity.

Step 4: Optimise and check

Differentiate:

At a non-vertical critical point,

Substitute \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\):

Multiplying by \(x+2y\) gives

Therefore

These are relationships between the variables at the critical points. We now return to the original constraint from Step 2.

If \(y=x\), then

giving \((\sqrt7,\sqrt7)\) and \((-\sqrt7,-\sqrt7)\), with

If \(y=-x\), then

giving \((\sqrt{21},-\sqrt{21})\) and \((-\sqrt{21},\sqrt{21})\), with

For completeness, Route B written as \(y=y(x)\) may miss points with a vertical tangent. These satisfy \(x+2y=0\). Substitution into the constraint gives \(x^2=28\), and hence \(D^2=35\), which lies between \(14\) and \(42\). Because the curve is closed and bounded, comparison of all candidates gives the absolute minimum and maximum.

Answer: Closest: \((\sqrt7,\sqrt7)\), \((-\sqrt7,-\sqrt7)\). Furthest: \((\sqrt{21},-\sqrt{21})\), \((-\sqrt{21},\sqrt{21})\).

Worked example 5: Maximise the volume of a cone

Question. A right circular cone has fixed slant height \(5\) cm. Its radius is \(r\) cm and its perpendicular height is \(h\) cm.

The dimensions satisfy the implicit constraint

Find the radius and height of the cone that give the maximum possible volume.

Solution

Step 1: Identify the target and constraint

The target quantity is the volume of the cone. The constraint is the fixed slant height, which links \(r\) and \(h\) through

Step 2: Write the model

The volume of a cone is

The separate constraint is

with \(r\ge 0\) and \(h\ge 0\).

Step 3: Use the constraint

We use Route B. Rather than writing \(h=\sqrt{25-r^2}\), we keep the constraint implicit and regard \(h=h(r)\). Differentiate the constraint with respect to \(r\):

Hence

Along the constraint, the target is the one-variable function

In Step 4, this expression for \(\frac{dh}{dr}\) will be substituted into \(\frac{dV}{dr}\), eliminating \(\frac{dh}{dr}\) from the derivative of the target quantity.

Step 4: Optimise and check

Differentiate:

Substitute \(\frac{dh}{dr}=-\frac{r}{h}\):

At an interior critical point,

Since \(r>0\) and \(h>0\) for a non-degenerate cone, multiply by \(h\) and factor out \(r\):

This gives a relationship between the variables. We now return to the original constraint from Step 2:

Therefore

and

At the endpoints \((r,h)=(0,5)\) and \((5,0)\), the volume is \(0\). Hence the interior critical point gives the absolute maximum.

The maximum volume is

Answer: The cone of maximum volume has radius \(\frac{5\sqrt6}{3}\) cm and height \(\frac{5}{\sqrt3}\) cm.