Find the maximum area of a triangle

Question

Find the maximum area of a triangle formed in the first quadrant by the x-axis, y-axis and a tangent line to the graph of f=(x+8)^-2. Area = ?

Answer 1

Equation of tangent line to the curve y = (x + 8)^-2 at point (x, (x + 8)^-2): y' = - 2(x + 8)^-3

Y = (x + 8)^-2 - 2(x + 8)^-3(X - x);

X = 0; Y = (x + 8)^-2+ 2x(x + 8)^-3 = (x + 8)^-3(3x + 8)

Y = 0; x + 8 = 2(X - x); X = 3x/2 + 4;

Area A(x) = XY/2 = (3x/2 + 4)(3x + 8)(x + 8)^-3/2 = (3x + 8)²(x + 8)^-3/4;

A'(x) = 1/4[6(3x + 8)(x + 8)^-3 - 3(3x + 8)²(x + 8)^-4] = 1/4(x + 8)^-4(3x + 8)(24 - 3x) = 0

Because x > 0 we have 3x - 24 = 0; x = 8; A'(x) at x =8 change sign from + to -, so we have max

max A9x) = A(8) = (24 + 8)²(16^-3/4 = 0.0625