y − 2 = m(x − 1) with intercepts as indicated in the figure.
The area A of the triangle is given by
.
Then and equals 0 when m = ±2; m must be negative.
(C) Let q = (x − 6)^2 + y^2 be the quantity to be minimized. Then
q = (x − 6)^2 + (x^2 − 4);
q′ = 0 when x = 3. Note that it suffices to minimize the square of the
distance.
(E) Minimize, if possible, xy, where x^2 + y^2 = 200 (x, y > 0). The
derivative of the product is , which equals 0 for x = 10. The
derivative is positive to the left of that point and negative to the right,
showing that x = 10 yields a maximum product. No minimum exists.
(C) Minimize . Since
,
