Solving a Classic IB Optimisation Problem Step by Step

Welcome, future mathematicians! If you're tackling IB AA HL or A-Level Mathematics, you know that optimisation problems are a cornerstone of the calculus curriculum. They can seem tricky at first, blending geometry, algebra, and differentiation into one neat package. But once you crack the method, they become incredibly satisfying to solve.

Today, we're going to walk through a classic example: finding the maximum possible area of a rectangle inscribed in a circle. We'll break it down step-by-step, just like a tutor would, with clear explanations and diagrams. Let's dive in!

Problem Setup

Here's our task: A rectangle is inscribed in a circle of radius r. Find the dimensions of the rectangle that will give the maximum possible area.

First, let's visualise the problem. A picture is worth a thousand words, especially in maths. We have a circle, and inside it, a rectangle whose four corners touch the circle's circumference.

r θ x y

To solve this, we need to create a function that describes the area of the rectangle. We can then use calculus to find the value that maximises this function.

Setting Up the Function

To find the area of the rectangle, we need its width and height. Let's place the centre of the circle at the origin (0,0) of a Cartesian coordinate system. The vertices of the rectangle will then have coordinates (x, y), (-x, y), (-x, -y), and (x, -y).

The width of the rectangle is 2x and the height is 2y. So, the area A is:

$$A = (2x)(2y) = 4xy$$

This is great, but we have a problem: our function A depends on two variables, x and y. To perform differentiation, we need to express A in terms of a single variable. This is where our constraint comes in — the rectangle is inscribed in a circle of radius r.

The equation of a circle centered at the origin is $x^2 + y^2 = r^2$. We can use this to express y in terms of x:

$$y = \sqrt{r^2 - x^2}$$

Now, we can substitute this back into our area formula:

$$A(x) = 4x\sqrt{r^2 - x^2}$$

Perfect! We now have the area A as a function of a single variable, x. Note that the domain for x is 0 < x < r.

Alternative approach using trigonometry:
We could also define the vertices using an angle θ, as shown in the diagram. Here, x = r cos(θ) and y = r sin(θ). The area would be $A(\theta) = 4(r \cos(\theta))(r \sin(\theta)) = 2r^2\sin(2\theta)$. This is often a more elegant way to solve the problem! Today, we'll stick to the Cartesian approach as it's more common in introductory calculus.

Differentiation & Critical Points

To find the value of x that maximises the area, we need to find the derivative of A(x) with respect to x and set it to zero. This will give us the critical points.

Our function is $A(x) = 4x(r^2 - x^2)^{1/2}$. We'll use the product rule [d/dx(uv) = u(dv/dx) + v(du/dx)] and the chain rule.

Let u = 4x and $v = (r^2 - x^2)^{1/2}$.

  • du/dx = 4
  • $\frac{dv}{dx} = \frac{1}{2}(r^2 - x^2)^{-1/2} \times (-2x) = -\frac{x}{\sqrt{r^2 - x^2}}$

Now, let's apply the product rule:

$$A'(x) = 4x \cdot \left[-\frac{x}{\sqrt{r^2 - x^2}}\right] + 4\sqrt{r^2 - x^2}$$

$$A'(x) = -\frac{4x^2}{\sqrt{r^2 - x^2}} + 4\sqrt{r^2 - x^2}$$

To solve for A'(x) = 0, we can combine these terms by finding a common denominator:

$$A'(x) = \frac{-4x^2 + 4(r^2 - x^2)}{\sqrt{r^2 - x^2}}$$

$$A'(x) = \frac{4r^2 - 8x^2}{\sqrt{r^2 - x^2}}$$

Now, we set the numerator to zero to find the critical points:

$$4r^2 - 8x^2 = 0$$

$$8x^2 = 4r^2$$

$$x^2 = \frac{r^2}{2}$$

$$x = \frac{r}{\sqrt{2}}$$

(We take the positive root since x represents a physical length).

So, our only critical point in the domain (0, r) is $x = r/\sqrt{2}$.

Second Derivative Test

We've found a critical point, but is it a maximum, a minimum, or a point of inflection? To be sure, we can use the Second Derivative Test. If A''(x) < 0 at our critical point, then we have a local maximum.

Let's find the second derivative, A''(x). We'll differentiate $A'(x) = \frac{4r^2 - 8x^2}{(r^2 - x^2)^{1/2}}$ using the quotient rule.

It can get a bit messy, but there's a simpler way for this specific problem. Look at the numerator of A'(x): $g(x) = 4r^2 - 8x^2$. Since the denominator $\sqrt{r^2 - x^2}$ is always positive in our domain, the sign of A'(x) is determined entirely by the sign of g(x).

  • When $x < r/\sqrt{2}$, $8x^2 < 4r^2$, so $A'(x)$ is positive (the area function is increasing).
  • When $x > r/\sqrt{2}$, $8x^2 > 4r^2$, so $A'(x)$ is negative (the area function is decreasing).

Since the derivative changes from positive to negative at $x = r/\sqrt{2}$, this confirms that we have found a local maximum. This is often called the First Derivative Test and avoids a more complicated differentiation.

Final Answer & Insight

We found the value of x that maximises the area. Now we need to find the dimensions of the rectangle and the maximum area itself.

The value for x is:

$$x = \frac{r}{\sqrt{2}}$$

Now let's find the corresponding value for y:

$$y = \sqrt{r^2 - x^2} = \sqrt{r^2 - \frac{r^2}{2}} = \sqrt{\frac{r^2}{2}} = \frac{r}{\sqrt{2}}$$

Wait a minute. We found that x = y. This means the width of the rectangle (2x) is equal to its height (2y). A rectangle with equal sides is... a square!

The dimensions of the rectangle with maximum area are:

  • Width = $2x = 2(r/\sqrt{2}) = r\sqrt{2}$
  • Height = $2y = 2(r/\sqrt{2}) = r\sqrt{2}$

And the maximum area is:

$$A = (r\sqrt{2}) \times (r\sqrt{2}) = 2r^2$$

The beautiful insight here is that for many symmetrical optimisation problems, the optimal solution is also symmetrical. The most "area-efficient" rectangle you can fit inside a circle is a perfect square.

Conclusion

We did it! By setting up a function for the area, finding its derivative, identifying the critical point, and testing for a maximum, we solved this classic optimisation problem. The key steps were translating a geometric problem into an algebraic function and then using the power of calculus to find the peak of that function.

These problems aren't just abstract exercises; they form the basis of finding optimal solutions in engineering, finance, and science. Mastering them is a big step forward in your mathematical journey.

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