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+6)^−2

Question

I would appreciate some assistance on this question

Answer 1

I cannot think of any other approach than the straightforward brute force,

which is messy and error prone.

So make sure to check this for mistakes.

f(x) = (x+6)^-2

f'(x) = -2(x+6)^-3

Any point on the function f is of the form (a, (a+6)^-2) for some a.

The tangent line through that point will have the slope f'(a).

Therefore the tangent has the equation

y = f'(a)x - f'(a)a + f(a)

= -2(a+6)^-3 x + 2(a+6)^-3 a + (a+6)^-2

Denote

h(a) = 2(a+6)^-3 a + (a+6)^-2 be the y-intercept of the tangent,

w(a) = (2(a+6)^-3 a + (a+6)^-2)/2(a+6)^-3 be the x-intercept of the tangent,

A(a) = h(a)w(a)/2 be the area of the triangle formed by the tangent.

We are to maximize A(a) as the value a ranges over the domain [0, ∞).

Manipulating

w(a) = (2(a+6)^-3a + (a+6)^-2)/2(a+6)^-3

= (2a + a + 6)/2

= (3a + 6)/2

A(a) = h(a)w(a)/2

= (2(a+6)^-3a + (a+6)^-2)(3a + 6)/2

= (a+6)^-2(2(a+6)^-1a + 1)(3a + 6)/2

= (a+6)^-2(2a/(a+6) + 1)(3a + 6)/2

= (a+6)^-2((3a+6)/(a+6))(3a + 6)/2

= (a+6)^-3(3a+6)²/2

A'(a) = (-3(a+6)^-4(3a+6)² + 2(a+6)^-3(3a+6))/2

= (-3(3a + 6) + 2(a+6)) (a+6)^-4 (a+6)/2

=(-7a - 6) (a+6)^-4(a+6)/2

A' is negative in the domain [0, ∞).

Therefore the maximum occur at the boundary point

A(0) = (6^-3 × 6²)/2 = 1/12