SECTION 3.5 Quadratic Discriminant 165
Example 4. Given that x^2 + 2y^2 = 6, find the maximum value of
x+y.
Solution. This problem appears quite a bit different (and more dif-
ficult) than the preceding examples, but it’s not, and it fits in very
well to the present discussion.^6 This problem is very geometrical
in nature, as the “constraint equation”x^2 + 2y^2 = 6 is an ellipse
and the graphs ofx+y=c(c= constant) are parallel lines (with
slope−1). We seek that value ofcwhich gives the maximum value
ofx+y. See the graphic below:− 6 − 4 − 2 2 4 6− 4− 3− 2− 11234xyD>0D=0x +2y =6 D<0(^22)
parallel lines
x+y=c
Equation 1: x²+2y²=6Equation 2: x+y=− 2
Clearly the maximum value ofEquation 3: x+y=4.5Equation 4: x+y=a x+y will occur where this line is
tangent to the ellipse. There will be two points of tangency, one
in the third quadrant (where a minimum value ofx+ywill occur)
and one in the first quadrant (where the maximum value ofx+y
occurs). Next, if we solvex+y=cforyand substitute this into
x^2 + 2y=6, then a quadratic equation inxwill result. For tangency
to occur, one must have that the discriminant is 0. Fromy=c−x,
obtain
x^2 + 2(c−x)^2 −6 = 0 =⇒ 3 x^2 − 4 cx+ 2c^2 −6 = 0.
This leads to
0 =D= 16c^2 −12(2c^2 −6) =− 8 c^2 + 72 =⇒c=± 3.
(^6) Problems of this sort are often not considered until such courses as Calculus III, where the
method of Lagrange multipliers is applied.