Among all pairs of numbers (x,y) such that 4x+2y=18, find the pair for which the sum of squares, x^2+y^2, is minimum.

Question

Among all pairs of numbers (x,y) such that 4x+2y=18, find the pair for which the sum of squares, x^2+y^2, is minimum.

Answer 1

The answer lies in the category, "Min/Max Applications Of Quadratic Functions"

 

First, rearrange the first equation to solve for y in terms of x. 

Subtract 4x from sides, and then divide both sides by 2.  You then have  y  =  -2x + 9.

Substitute the right side of this equation in place of y in the second expression.

You end up with   x^2 + y^2  =  x^2 + (-2x + 9)^2  =  x^2  +  4x^2  -  36x  + 81  =  5x^2  -  36x  =  81.

 

 

You can find the minimum of the function  f(x)  =  5x^2  -  36x  +  81 by finding its derivative (slope function), setting that derivative equal to zero, and solving that equation for x.  The derivative function, f'(x), equals 10x - 36.

Solving 0  =  10x  - 36 for x yields  x = 3.6.

 

With x  =  3.6, the first equation yields  y  =  -7.2  +  9  =  1.8, and the second expression, x^2 + y^2, yields a minimum value of 16.20.