Question

Among all pairs of numbers (x,y) such that 3x+y=15, find the pair for which the sum of squares, x^2+y^2, is minimum. Write your answers as fractions reduced to lowest terms.

Answer 1

Answer:

Minimum value occurs at (x, y) = (-5,0)

Step-by-step explanation:

Given: 3x + y =15

Rewriting this equation to Slope - Intercept form, we get:

y = -3x + 15

Squaring both sides,

$y^2 = (-3x + 15)^2$

$\implies y^2 = 9x^2 - 90x + 225$

Substituting in $x^2 + y^2,$ we get:

$x^2 + 9x^2 - 90x + 225$

= $10x^2 - 90x + 225$

Comparing this equation with $ax^2 + bx + c$ we get $a =10$ and $b = - 90$.

The minimum value occurs at the axis of symmetry given by $x = \frac{-b}{2a}$.

Therefore, x = - $\frac{-90}{2.10}$

$\implies x = 5$

If x = 5, y = -3(5) + 15 = 0

Thus, the point becomes (5,0) where it is minimum.