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.
On
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 $120/17$.
On
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 $$\sqrt{x_0^2+y_0^2}=\frac{120}{17^2}\sqrt{15^2+8^2\strut}=\frac{120}{17}.$$
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 ${1 \over 2}{2\,\cdot\,\text{area of the triangle} \over \sqrt{15^2+8^2}} = {15 \cdot 8 \over \sqrt{225+64}} = {120 \over \sqrt{289}} = {120 \over 17}$.