You may also like

problem icon

Folium of Descartes

Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.

problem icon

Witch of Agnesi

Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.

problem icon

Quick Route

What is the quickest route across a ploughed field when your speed around the edge is greater?

Least of All

Stage: 5 Challenge Level: Challenge Level:1

Thank you Ruth from Manchester High School for Girls for your solution to this problem.

Most people's first reaction to this question is exactly the same as Ruth's:

"My initial conjecture was that the minimum value of $f(x)$ is when $x=0$ for any value of $a$, because the function is even and increases as $|x|$ increases."

Using calculus, we shall see that this is not so and that the minimum value of $f(x)$ does depend on the value of $a$ .

The given expression is \begin{eqnarray} f(x) &=& (1 + (a+x)^2)(1 + (a-x)^2)\\ &=&(1 + x^2 + a^2 +2ax)(1 + x^2 +a^2 -2ax)\\ &=& (1 + x^2 + a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 + a^2) + (1 +a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 - a^2) + (1+a^2)^2. \end{eqnarray}
As this is a quartic in $x$ there will be one or three turning points.

Differentiating $f$ to find the minima:
$$f'(x) = 4x^3 + 4x(1-a^2) = 4x(x^2 + (1 - a^2))$$

Case 1 : $(1 - a^2) < 0$.
The derivative $f'(x) = 0$ for $x = 0$ and $x = \pm \sqrt (a^2 - 1)$. The second derivative $f''(x) = 12x^2 + 4(1 - a^2) < 0$ at $x = 0$ which gives a maximum value but $f''(x) = 12x^2 + 4(1 - a^2) > 0$ for $x = \pm \sqrt (a^2 - 1)$ giving two minimum points on the quartic where $x = \pm \sqrt (a^2 - 1)$. The minimum value at each point is $f(x) = 4a^2$ where the position of X for these minimum values is clearly dependent of $a$.

Case 2 : $(1 - a^2) > 0$.
The derivative $f'(x) = 0$ if and only if $x = 0$. The second derivative
$f''(x) = 12x^2 + 4(1 - a^2) > 0$ so there is one minimum value $f(x)= (1+a^2)^2$ where the position of X, at $x=0$, is independent of $a$ agreeing with the conjecture.

Case 3 : $(1 - a^2) = 0$.
Note that where $a^2 = 1$ there is a single minimum $f(x) = 4$ at $x=0$ giving continuity between Case 1 and Case 2.

The most likely first conjecture agrees with Case 2 but does not allow the possibility of Case 1. Realising that the function we are minimizing is a quartic we should have taken into account the possibility of two minimum values.