Find the point on the hyperbola xy=8 that is closet to the point (3,0)

Use distance formula to get a distance function in terms of x representing the distance from a point on curve (x,y) to point (3,0)



y = 8/x


Minimize the function by setting derivative equal to 0.

Simplify and solve for x

This becomes a 4th degree polynomial, you can use graphing calculator to find zeros.
Also intuitively you can recognize that 64 = 4^3, so if x=4 you get 
1 = 1

Therefore x=4 is a solution, substituting give  y =2

I’m not absolutely sure but i think it’s 7x6x5x4x3x2=5040 because as the first character you can take any of the 7 given then you are left with 5 then 6 then 4 etc.

The operation is suck a c@ck? !