Maximum and Minimum problems

[Tinkermom]

Suppose that x + y = sqrt2. Find the minimum value of the quantity x^2 + y^2.

I took x + y = sqrt2 and solved for y. I got y = sqrt2 - x.

Do I then take x^2 + (sqrt2 - x)^2 = 0 and solve for x? If so, then what do I do after that?

Thanks for your help.

 

[mmm4444bot]

Do I then take x^2 + (sqrt2 - x)^2 = 0 and solve for x?

Not quite. We don't set x^2 + y^2 equal to zero because we don't know its value.

We're trying to find its smallest value, and that value might not be zero.

Expand the expression x^2 + (sqrt[2] - x)^2 by multiplying out everything and combining like-terms.

You'll end up with a quadratic polynomial in the form Ax^2 + Bx + C.

The smallest value of this polynomial occurs at the vertex of its graph.

Use the formula below (in blue) to determine the x-coordinate at the vertex, and then use that value of x to evaluate the quadratic polynomial. That gives you the corresponding y-coordinate, at the vertex.

It's the y-coordinate of the vertex that you're looking for because that's the smallest value.

x-coordinate of vertex point = -B/(2A)

In other words, the x-coordinate of a parabola's vertex is what's left in the Quadratic Formula when the discriminant (i.e., the radicand B^2-4AC) is zero.

I welcome specific questions.

Cheers ~ Mark