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[SOLVED] 3.1.009 AP Calculus Exam shortest distance between curve and origin


Well-known member
Jan 31, 2012
Find the x-coordinate of the point on $f(x)=\dfrac{4}{\sqrt{x}}$
that is closest to the origin.

a. $1$
b. $2$
c $\sqrt{2}$
d $2\sqrt{2}$
e $\sqrt[3]{2}$

not real sure but, this appears to be dx and slope problem
I thot there was an equation for shortest distance
between a point and a curve but couldn't find it offhand


Well-known member
MHB Math Helper
Mar 1, 2012
$D^2 = (x-0)^2 + \left(\dfrac{4}{\sqrt{x}} - 0\right)^2 = x^2 + \dfrac{16}{x}$

minimizing $D^2$ will minimize $D$ ...

$\dfrac{d(D^2)}{dx} = 2x - \dfrac{16}{x^2} = 0$

finish and confirm the value is a minimum


Staff member
Feb 24, 2012
Another approach would be Lagrange Multipliers (optimization with constraint. The objective function could be the square of the distance:

\(\displaystyle f(x,y)=x^2+y^2\)

Subject to the constraint:

\(\displaystyle g(x,y)=y-\frac{4}{\sqrt{x}}=0\)


\(\displaystyle 2x=\lambda\left(2x^{-\frac{3}{2}}\right)\)

\(\displaystyle 2y=\lambda(1)\)

This implies:

\(\displaystyle y=\frac{x^{\frac{5}{2}}}{2}\)

Substituting into the constraint, there results:

\(\displaystyle \frac{x^{\frac{5}{2}}}{2}-\frac{4}{\sqrt{x}}=0\)

This leads to the same root as above, and to verify it is a miniimum we could pick another point on the constraint to verify the objective function is greater at that point than at our critical point.


Well-known member
Jan 31, 2012

I've never did anything with Lagrange


Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
My first thought would be to just try each possibility:
a) x= 1. The point is (1, 4) which has distance $\sqrt{17}$, about 4.12 from the origin.
b) x= 2. The point is (2, 4/\sqrt{2}) which has distance $\sqrt{4+ 8}= \sqrt{12}$, about 3.46, from the origin.
c) $x= \sqrt{2}$. The point is $(\sqrt{2}, 4/\sqrt[4]{2})$ which has distance $\sqrt{2+ 16/\sqrt{2}}= \sqrt{2+ 8\sqrt{2}}$, which is about 3.65, from the origin.
d) $x= 2\sqrt{2}$. The point is $(2\sqrt{2}, 4/\sqrt[4]{8})$ which has distance $\sqrt{8+ 16/\sqrt{8}}= \sqrt{8+ 8/\sqrt{2}}= \sqrt{8+ 4\sqrt{2}}$, which is about 3.70 from the origin.
e) $x= \sqrt[3]{2}$. The point is $\sqrt[3]{2}, 4/\sqrt[6]{2})$ which distance $\sqrt{\sqrt[3]{4}+ 16/\sqrt[3]{2}}$ which is about 3.78 from the origin.

Of the four distances, the smallest is 3.46 so (b) x= 2 gives the point closest to the origin!

Heavy use of calculator, light use of brain!