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+3)^-2

^Area=

I don't understand what I'm supposed to do with the function to figure this one out.

Thank you for any help!

Answer 1

Let (x₀, y₀) is tangent point. Then slope of tangent line is m = y' at x = x_0, m = -2(x₀ + 3)^-3 and equation of tangent line is y = (x₀ + 3)^-2 - 2(x₀ + 3)^-3(x - x₀).

Now y-intercept is (x₀ + 3)^-2 + 2(x₀ + 3)^-3x₀; x-intersept is (x₀ +3)/2 + x₀ = 3/2(x₀ + 1);

Area of triangle A = 1/2·3/2(x₀ + 1)(x₀ + 3)^-33(x₀ + 1) = 9/4(x₀ + 1)²(x₀ + 3)^-3;

We get area as function of x₀

Naw A' = .9/4[2(x₀ + 1)(x₀ + 3)^-3 - 3(x₀ + 1)²(x₀ + 3)^-4] = 9/4 (x₀ + 1)(x₀ + 3)^-4(2x₀ + 6 - 3x₀ - 3) =

9/4(x₀ + 1)(x₀ + 3)^-4(3 - x₀) = 0. In first quadrant we have only 1 critical point x₀ = 3.

Now left frm x₀ = 3 A' > 0 and right from x₀ = 3 A' < 0. So at x₀ = 3 area A is max

A_max = 9/4(3 + 1)²(3 + 3)^-3 = 9/4·16·1/6³ = 1/6

Answer 2

Graph the function, try a few number and see where the area seems to change, it gives

clues at least where the solution might be.

tangent line would have slope = derivative of the function

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

If x= 1/2 then y = 1/(3.5)^2 = 1/12.25

and the slope of the tangent line is

-2(3.5)^-3 = about -2/43

plug those values into the point slope formula to get the tangent line's equation

y-1/12.25 = -2/43(x-1/2) = -2x/43 - 3.5/43

y = -2x/43 - 1/43 + 3.5/43 = -2x/43 + 2.5/43

the tangent line has a y-intercept of 2.5/43 = about .06

the x-intercept is when y=0

0=-2x/43 + .262

2x = 43(.262)

x = 43(.131) = 5.633

Area of the triangle is (1/2)(x-intercept)(y-intercept) = (1/2)(5.633)(.06) = .,16 = about 4/25 = about 1/6

Try x= 1 and x= 1/4 or x= 1/3, they seem to give smaller areas.

x= 1 leads to Area = 9/64 or about 1/8

x=1/4 leads to area = 30/169 or between 1/5 or 1/6

anything larger than x= 1 gives a decreasing area approaching zero as x approaches infinity

Max Area = about 30/169 > 1/6 but < 1/5

the function has a vertical asymptote at x=-3

y is always positive

it also has a horizontal asymptote at y=0

It has 2 branches with a discontinuity at x=-3