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)
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
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
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
To 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), we can follow these steps:
Step 1: Understand the problem
The problem asks for the maximum area of a triangle formed by three specific entities - the x-axis, y-axis, and a tangent line. The equation y = (x+3)^(-2) represents a curve, and we need to find the tangent line that touches it in the first quadrant.
Step 2: Find the derivative
To find the equation of the tangent line, we need to find the derivative of the given curve. Differentiating y = (x+3)^(-2) with respect to x will give us the slope of the tangent line at any point on the curve.
dy/dx = -2(x+3)^(-3) * 1
= -2/(x+3)^3
Step 3: Find the x-coordinate of the tangent point
To find the x-coordinate of the point where the tangent line touches the curve, we need to set the derivative equal to zero and solve for x. This will give us the critical point, where the slope of the tangent line is zero.
-2/(x+3)^3 = 0
Since the numerator is zero, the only possible solution is when the denominator is zero.
x+3 = 0
x = -3
So, the tangent point on the curve is when x = -3.
Step 4: Find the y-coordinate of the tangent point
Substitute the x-coordinate (-3) into the equation y = (x+3)^(-2) to find the y-coordinate of the tangent point.
y = (-3+3)^(-2)
= 0^(-2)
= 0
So, the y-coordinate of the tangent point is 0.
Step 5: Find the equation of the tangent line
Now that we have the x and y coordinates of the tangent point, we can find the equation of the tangent line using the point-slope form.
Using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope, we can substitute the values we found in Step 3 and Step 4.
y - 0 = -2/(x+3)^3 * (x - (-3))
y = -2/(x+3)^3 * (x+3)
Simplifying this equation will give us the equation of the tangent line.
Step 6: Find the length of the base of the triangle
The length of the base of the triangle is equal to the x-coordinate of the tangent point, which we found to be x = -3.
Step 7: Find the height of the triangle
The height of the triangle is equal to the y-coordinate of the tangent point, which we found to be y = 0.
Step 8: Calculate the area of the triangle
Using the formula for the area of a triangle (Area = 1/2 * base * height), we can substitute the values we found in Step 6 and Step 7.
Area = 1/2 * (-3) * 0
Area = 0
Therefore, the maximum area of the triangle formed in the first quadrant by the x-axis, y-axis, and a tangent line to the graph of y = (x+3)^(-2) is 0.