Animated Solution for Mathematics - Trigonometry: If $p$ and $q$ are positive real numbers such that $p^2+q^2=1$, then the maximum value of $(p+q)$ is

Imagine you are standing on a coordinate plane with the constraint $p^2+q^2=1$. If you have ever studied geometry, your mind should immediately jump to the image of a beautiful, perfect unit circle centered at the origin.

However, the problem specifies that $p$ and $q$ are strictly positive. This means we are not looking at the entire circle, but only the elegant arc residing in the first quadrant.

We are tasked with maximizing the sum $f(p,q)=p+q$. Let us call this sum $k$. If we rearrange this, we get $q=-p+k$, which is the equation of a straight line with a slope of $-1$.

As we vary $k$, we are essentially sliding this line across the plane. Our goal is to find the largest possible $k$ such that this line still touches our arc.

While the geometric approach is visually stunning, sometimes we need a more surgical tool. This is where the magic of parametric coordinates comes into play.

Since any point on the unit circle satisfies $p^2+q^2=1$, we can describe these points using the most fundamental trigonometric functions. Let us set $p=\cos \theta$ and $q=\sin \theta$.

Because $p$ and $q$ are positive, our angle $\theta$ must be strictly between $0$ and $\frac{\pi }{2}$. By making this substitution, we have transformed a two-variable problem into a single-variable function:

A Mathematical Shortcut

Now, we are looking for the maximum value of $\cos \theta +\sin \theta$. This is a specific instance of the general form $a\cos \theta +b\sin \theta$.

There is a well-known, powerful result in trigonometry that states the range of this expression is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$. This formula is a lifesaver in JEE problems.

By mapping our expression to this form, we identify $a=1$ and $b=1$. Plugging these into our formula, the maximum value becomes:

The Moment of Tangency

Let us pause and reflect on what we have just found. We calculated that the maximum value is $\sqrt{2}$.

This occurs when $\theta =\frac{\pi }{4}$, or $45^{\circ }$. At this angle, the coordinates are:

If you look back at our geometric intuition, this is the exact point where the line $p+q=\sqrt{2}$ is tangent to the circle. It is the point of perfect symmetry, where the line kisses the arc.

The math and the geometry are in perfect harmony. You have successfully navigated the constraints, applied the right algebraic tool, and verified the result with geometric intuition. The final maximum value is $\sqrt{2}$.