0
$\begingroup$

How would you find the minimum value of the equation above without calculus or graphing? Since the lowest value of a square root is zero, I set the equation above to zero but ended up getting a negative discriminant.

$\endgroup$
2
  • $\begingroup$ Try putting x = 128 $\endgroup$ – Shailesh Dec 2 '19 at 3:33
  • 1
    $\begingroup$ use the axis of symmetry equation given by $x=\frac{-b}{2a}$. This gives the minimum value of your quadratic which is also the minimum value of your function. If its negative, ignore it and note that your quadratic approaches positive infinity so it will be equal to zero at some point. (which means your min value is zero). Otherwise, that $x$ value will give you the vertex which is also the minimum. $\endgroup$ – Sina Babaei Zadeh Dec 2 '19 at 3:43
5
$\begingroup$

Well, let's begin by rewriting it. Note that $\sqrt{50x^2-12800x+820000}$ is minimized when $50x^2-12800x+820000$ is minimized*, so we'll temporarily ignore the square root.

We have $$50x^2-12800x+820000$$ $$=50(x^2-256x+128^2)+(820,000-50(128)^2)$$ $$=50(x-128)^2+800$$

The minimum value of this quadratic, therefore, is $800$.

And so, the minimum value of $\sqrt{50x^2-12800x+820000}$ is $\sqrt{800}=\boxed{20\sqrt{2}}$

*It is only minimized if this quadratic is positive for all real $x$. Otherwise, the minimum, as you suggested, would be $0$.

$\endgroup$
1
$\begingroup$

The vertex of the parabola $y = ax^2 + bc + c$ occurs at $x = -b/2a = 128$. Substitute that value for $x$, find $y$ and take the square root. Factoring out $50$ and paying attention to $b=128$ will save you some arithmetic.

$\endgroup$
2
  • $\begingroup$ i'm literally dumb for not thinking of that +1 $\endgroup$ – Saketh Malyala Dec 2 '19 at 3:47
  • 1
    $\begingroup$ @SakethMalyala No, not dumb at all. Many math problems look easy after you see a solution. $\endgroup$ – Ethan Bolker Dec 2 '19 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.