Find the maximum area of a triangle formed in the first quadrant by the 𝑥 x -axis, 𝑦 y -axis and a tangent line to the graph of 𝑓=(𝑥+4)−2 f = ( x + 4 ) − 2 .

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+4)^−2

Answer 1

Let (x₀, y₀) is tangent point. Then f'(x) = - 2(x + 4)^-3; f'(x₀) = - 2(x₀ + 4)^-3. Equation of tangent line:

y = (x₀ + 4)^-2 - 2(x₀ + 4)^-3(x - x₀);

y- intercept: x = 0 y = (x₀ + 4)^-2 + 2x₀(x₀ + 4)^-3 = (x₀ + 4)^-3(3x₀ + 4).

x- intercept: y = 0; x = (3x₀ + 4)/2

Area of triangle A = (3x₀ + 4)²(x₀ + 4)^-3/4 =1/4(3x₀ + 4)²/(x₀ + 4)³.

dA/dx₀ = 1/4[6(3x₀ + 4)(x₀ + 4)³ - (3x₀ + 4)²·3(x₀ + 4)²]/(x₀ + 4)⁶ = 1/4(3x₀ + 4)(6x₀ + 24 - 9x₀ - 12)/(x₀ + 4)⁴ =

1/4((3x₀ + 4)(12 - 3x₀)/((x₀ + 4)⁴ = 0; x₀ = - 3/4(not 1st quadrant) or x₀ = 4.

At x = 4 area A has maximum , because dA/dx changes sign from plus to minus from left to right.

Max A = A(4) = 1/4(12 + 4)²/(4 + 4)³ = 0.125