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 𝑦=(𝑥+3)^(−2)

Answer 1

dy/dx = -2(x+3)^-3 or -2/(x+3)^3

let P(a,1/(a+3)^2) be any point on the curve
then at P, the slope of the tangent is -2/(a+3)^3
and we can write the equation of the tangent as
y - 1/(a+3)^2 = (-2/(a+3)^3 (x - a)

I want to find the x and y intercepts of this tangent
let x = 0 , then y = 1/(a+3)^2 + 2a/(a+3)^3
y = (3a+3)/(a+3)^3

if y = 0, then - 1/(a+3)^2 = -2x/(a+3)^3 +2a/(a+3)^3
2x/(a+3)^3 = 2a/(a+3)^3 + 1/(a+3)^2
multiply each term by (a+3)^3

2x = 2a + a+3
x = (3a+3)/2

area = (1/2)xy
= (1/2)*(3a+3)/2 *(3a+3)/(a+3)^3
= (9/2)(a+1)^2 / (a+3)^3

ok, your turn.
Find d(area)/da using the quotient rule, set that numerator equal to zero, and solve for a
Once you have a, find x and y, and thus the area

check my algebra, but my a value came out to be a whole number and the
maximum area a nice exact fraction.
Let me know what you get

Answer 2

thank you for your help!!!
here is my attempt to this

d(area)/da=(9/2)(a+1)(a+3)^2*(2(a+3)-3(a+1))/(a+3)^4
d(area)/da=9(a+1)(-a+3)/2(a+3)^4

9(a+1)(-a+3)/2(a+3)^4=0
a=-1 , 3

put a=3 into x = (3a+3)/2
x=6
put x=6 into y = (3a+3)/(a+3)^3
y=1/18

Area=1/2(6*1/18)=1/6

Answer 3

I got a = 3 , rejected the a = -1 since we wanted only the first quad

My area was different, check both mine and yours
If we go back to
area = (9/2)(a+1)^2 / (a+3)^3
area = (9/2)(4^2) / (6^3)
= (9/2)(16)/216
= 1/3