#
Problem
of the Week

Problem
E and Solution

Keeping
it Small

## Problem

Clara is learning to code, so she has written a program to help her
practice what she has learned so far.

Clara’s program takes two real numbers as input, called \(A\) and \(B\). First, her program doubles \(A\), squares the result, and then reduces
this result by four times \(A\). The
result is called \(C\). Then her
program squares \(B\), and then
increases this result by six times \(B\). The result is called \(D\). Finally, her program outputs the sum
of \(C\) and \(D\).

Determine the minimum possible output of Clara’s program, and the two
input values that produce this output.

## Solution

In order to minimize the final output, we need to minimize both \(C\) and \(D\).

First, let’s minimize \(C\). Clara’s
program doubles \(A\) to get \(2A\). It squares this result to get \((2A)^2 = 4A^2\). It then reduces this
number by \(4A\), to get \(C=4A^2 -4A\). Thus, we need to minimize
\(4A^2 -4A\). This is a quadratic and
so represents a parabola. Since the coefficient of \(A^2\) is positive, it opens up and so its
minimum value occurs at its vertex. We can find the vertex by completing
the square. $$\begin{align}
4A^2 - 4A &= 4(A^2 - A)\\
&= 4\left(A^2 - A + \frac{1}{4}-\frac{1}{4}\right)\\
&= 4\left(A^2 - A + \frac{1}{4}\right)-1\\
&= 4\left(A - \frac{1}{2}\right)^2-1
\end{align}$$ The vertex is at \(\left(\frac{1}{2}, -1\right)\), and so the
minimum value of \(C=4A^2 -4A\) is
\(-1\) and occurs when \(A=\frac{1}{2}\).

Now let’s minimize \(D\). Clara’s
program squares \(B\) to get \(B^2\). It then increases the result by
\(6B\) to get \(D=B^2+6B\). So we need to minimize \(B^2+6B\). This is a quadratic and so
represents a parabola. Since the coefficient of \(B^2\) is positive, it opens up and so its
minimum value occurs at its vertex. We can find the vertex by completing
the square. $$\begin{align}
B^2 +6B &= (B^2 +6B + 9) - 9\\
&= (B + 3)^2-9
\end{align}$$ The vertex is at \((-3,
-9)\), and so the minimum value of \(D=B^2+6B\) is \(-9\) and occurs when \(B=-3\).

Therefore, the minimum possible output of Clara’s program is \(-1 + (-9) = -10\) and occurs when \(A=\frac{1}{2}\) and \(B=-3\).

*Aside:* This problem is essentially asking us to minimize the
multivariable function \(f(x,y) = 4x^2 - 4x+
y^2 + 6y\).