About the Application of Cauchy-Schwarz (Basic): maximum of $3x+4y$ for $x^2+y^2 \leq 16$ Today I've seen a question like this:
$$\text{Given } x^2+y^2 \leq 16, \text{ what is the maximum value for } 3x+4y?$$.
What I've tried was the following:
$$3x+4y \leq \sqrt{x^2+y^2}\cdot\sqrt{9+16}$$
But the problem here is that I have to give a value for $\sqrt{x^2
+y^2}$ and I know that this would not give me a general solution.
What do you suggest?
 A: Hint. You proved that if $x^2+y^2 \leq 16$ then 
$$3x+4y \leq \sqrt{x^2+y^2}\cdot\sqrt{3^2+4^2}\leq \sqrt{16}\cdot \sqrt{25}=20.$$
Now try to find when the equality holds. Recall that in the Cauchy-Schwarz inequality, 
$$u\cdot v\leq \|u\|\cdot \|v\|$$
the equality holds iff $u=\lambda v$ for som $\lambda\in\mathbb{R}$. 
P.S. In your case let $u=(x,y)$ and $v=(3,4)$. Then $x=3\lambda$, $y=4\lambda$ and solve $16^2=x^2+y^2=(9+16)\lambda^2$. Thus $\lambda=\sqrt{16/25}=4/5$ ($\lambda=-4/5$ works as well). Hence equality holds for $(x,y)=(12/5,16/5)$.
A: You need to understand also, when the equality occurs.
It happens for $(3,4)||(x,y)$ and $x^2+y^2=16$, which says that $20$ is a maximal value.
A: This is also solvable using simple analytic geometry.
$x^2+y^2 \le 16$ is the interior of a disk with radius $4$, and $3x+4y=f$ is a down-sloping straight line for any given value of $f$.
We are looking for the maximal value of $f$ for lines that intersect this disk. Since $f$ is linear in both $x$ and $y$ it's obvious that the maximum must occur where the line is a tangent to the disk boundary. The intersection point must lie on the ray $(3t,4t)$ where $t$ is an arbitrary parameter, since this ray is perpendicular to every line defined by keeping $f$ constant.
So it boils down to solving $(3t)^2+(4t)^2=16$, which gives $t=\frac{4}{5}$, and from that follows $x=\frac{12}{5}$, $y=\frac{16}{5}$, and the maximum value is thus: $3x+4y=\frac{3 \times 12}{5} + \frac{4 \times 16}{5}=20$.
A: I know this is nothing about Cauchy-Schwarz, but maybe it helps.
The function $f(x,y)=3x+4y$ is harmonic in the disc $D_4(0)$. Now $f$ attains the maximum value on the boundary of $D_4(0)$ (see maximum principle) and this is given by the equation $x^2+y^2=4^2=16$.
