Expert: Alon Mandes - 11/29/2009

Question
We are on differentiation and derivative. This question is very confusing. I hope someone can help me as soon as possible. Thank you so much!

Let f(x)= 6-x^2
For 0 < w < squareroot of 6, let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point (w, 6 - w^2)

1. Find the equation of the tangent line when w =1
2. Find the x and y intercepts of your tangent line, and use those values to calculate A(1)
3. Find the equation of the tangent line at (w, 6-w^2) in terms of w, x and y
4. Use this tangent line to find an equation fr A(w)
5. Algebraically find the value of w that minimizes A(w) on the given domain. Jutify your answer

Answer
The tangent line has an equation from the form : y=mx+n . Let's find m & n :
m is the slope = f'(x) . Therefore m=f'(1). f'(x)=-2x --> f'(1)=-2 .
n is the intercept point of the line with the y-axis. To find it , we substitute the point
(1,6-1^2)=(1,5) : 5=-2*1+n --> n=7 . Therefore the tangent line equation y=-2x+7 .
Let's find the point of intercept of the line with the x-axis :
We substitute y=0 : 0=-2x+7 --> x=3.5
The Triangle formed is right one, thus the area will be : A=(3.5*7)/2=12.25

Let's find A, at point w :
m is the slope = f'(x) . Therefore m=f'(w). f'(x)=-2x --> f'(w)=-2w .
n is the intercept point of the line with the y-axis. To find it , we substitute the point
(w,6-w^2) : 6-w^2=-2w+n --> n=6-w^2+2w . Therefore the tangent line equation is:
y=-2wx+6-w^2+2w .
Let's find the point of intercept of the line with the x-axis :
We substitute y=0 : 0=-2wx+6-w^2+2w  --> x=(6-w^2+2w)/2w
The Triangle formed is right one, thus the area will be :
A(w)=[(6-w^2+2w)/2w]*(6-w^2+2w)/2=(1/4w)(6-w^2+2w)^2

To find minimum we derive & solve : A(w)=0 .

Alon.