Finding minimum value of $\sqrt{x^2+y^2}$

Question

Find the minimum value of $\sqrt{x^2+y^2}$, given $15x+8y=120$.

My attempt:

From $15x+8y=120$, I get $y=\frac{120-15x}{8}$. I substitute this value into $\sqrt{x^2+y^2}$, getting $\sqrt{\frac{289x^2-3600x+14400}{64}}$. I am stuck here as $289x^2-3600x+14400$ cannot be square-rooted.

For $a>0$ the minimum value of the parabola $y=ax^2+bx+c$ is attained at $x=-b/(2a)$ — Robert Z, Jan 18 '18 at 09:17
Sorry, I have no idea what that is too. Perhaps you could present a solution? Also, this worksheet from my teacher did not require the use of differentiation — QuIcKmAtHs, Jan 18 '18 at 09:17
The minimum value of $\sqrt{x^2+y^2}$ is attained at the same point as $x^2+y^2$ — egreg, Jan 18 '18 at 09:17
Hint: the minimum of the square root coincides with the minimum of the argument of the square root (provided it is non-negative), as the square root function is monotonous. — , Jan 18 '18 at 09:26
Max / min of $ y=a x^2 + bx+c, is , c- b^2/4a;,$ Plug in $(289,-3600, 14400 )$ and find its square-root — Narasimham, Jan 18 '18 at 16:02

score 1 · Answer 1 · answered Jan 18 '18 at 09:31

As you derived, $\sqrt{x^2+y^2}=\sqrt{\frac{289x^2-3600x+14400}{64}}$.

Note that the expression under the square root is a quadratic expression.

It is important to understand that when the number under the square root is smaller, the square root of the number is also smaller.

Therefore, we are interested in finding the minimum of the quadratic expression.

As you may know, the minimum value of a quadratic expression is $$c-\frac{b^2}{4a}$$. Therefore, the minimum of your quadratic expression is $14400/289$.

Thus the minimum value of the square root is $120/17$.

Artur Riazanov · Accepted Answer · 2018-01-18T09:31:29.007

This problem can be stated in a geometric way: find the distance from the point $(0,0)$ to the line $\{(x,y) \colon 15x + 8y = 120\}$. Let's find out where this line intersects with axis: $y=-{15 \over 8} x + {120 \over 15} = {-15 \over 8} x + 8$, so if $x=0$ then $y = 8$. On the other hand $x = {-8 \over 15} y + {120 \over 8} = {-8 \over 15} y + 15$, hence if $y=0$ then $x=15$. Therefore we need to find the altitude of the triangle with sides $15$ and $8$. Clearly it is ${1 \over 2}{2\,\cdot\,\text{area of the triangle} \over \sqrt{15^2+8^2}} = {15 \cdot 8 \over \sqrt{225+64}} = {120 \over \sqrt{289}} = {120 \over 17}$.

Wow, this is definitely a +1. I did not expect something so nice. — QuIcKmAtHs, Jan 18 '18 at 09:31

score 0 · Answer 3 · answered Jan 18 '18 at 11:14

Do some geometry first: this minimum is attained at the orthogonal projection of the origin on the straight line $(L)$ with equation $15x+8y=120$.

Now the vector $(15,8)$ is normal to this line. Thus we have to find the intersection of the line with parametric representation $\;x=15t,\; y=8t$ and $(L)$: $t$ must satisfy the equation $$15(15t)+8(8t)=120\iff t=\frac{120}{289}=\frac{120}{17^2}.$$ Thus $$\sqrt{x_0^2+y_0^2}=\frac{120}{17^2}\sqrt{15^2+8^2\strut}=\frac{120}{17}.$$

Finding minimum value of $\sqrt{x^2+y^2}$

3 Answers3