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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?

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  • $\begingroup$ Do you have access to the technique called "Lagrange multipliers"? (It's probably overkill, but it will get the job done.) $\endgroup$
    – Arthur
    Aug 30 '17 at 8:29
  • $\begingroup$ I have but ı am requested to solve it without lagrange:( $\endgroup$ Aug 30 '17 at 8:31
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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 $(x,y)=(12/5,16/5)$.

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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.

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  • $\begingroup$ I didn't think we could substitude a value to $x^2+y^2$ directly. If so ı am glad. Thank you:) $\endgroup$ Aug 30 '17 at 8:34
  • $\begingroup$ @Deniz Tuna Yalçın It follows from your work. Since $x^2+y^2\leq16$ , we obtain: $3x+4y\leq\sqrt{(3^2+4^2)(x^2+y^2)}\leq\sqrt{25\cdot16}=20.$ $\endgroup$ Aug 30 '17 at 8:37
  • $\begingroup$ This is Lagrange multipliers in a nutshell (exploiting the fact that we're on a circle centered at the origin to get $(3, 4)\|(x, y)$). It's not the full machinery, and I'm not saying it's wrong. $\endgroup$
    – Arthur
    Aug 30 '17 at 8:37
  • $\begingroup$ @Arthur It follows also from the equality case in C-S. I think LM is very ugly here. $\endgroup$ Aug 30 '17 at 8:39
  • $\begingroup$ @MichaelRozenberg Sure. It might be more correct to say that LM is a generalisation of this, rather than saying that this is a special case of LM. $\endgroup$
    – Arthur
    Aug 30 '17 at 8:40
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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: $3x+4y=\frac{3 \times 12}{5} + \frac{4 \times 16}{5}=20$.

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  • $\begingroup$ Wow! I never thought of it this way. Thank you:) $\endgroup$ Aug 30 '17 at 18:03
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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 $x^2+y^2=4^2=16$.

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