Expert: Paul Klarreich - 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
Questioner: Chloe
Country: United States
Category: Calculus
Private: No
Subject: Calculus derivative
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
......................................
Actually, they are telling you how to do the problem.  The only thing you need to know is the process for finding the tangent line, which is:

A. Find m = dy/dx at the point in question.
B. Use the point-slope form:  

  y - y0 = m(x - x0)
.......................................
So part 1 goes:

m = dy/dx = - 2x
at x0 = 1,  m = -2, and  y0 = 2 - 1 = 1.

So  

y - 1 = -2(x - 1)

y - 1 = -2x + 2

y = - 2x + 3.

Now the intercepts are:

y-int:  let x = 0,  y = 3
x-int:  let y = 0,  x = 3/2

Your triangle has b = 3/2, h = 3,  A = 9/4
...............................
Now for any w, do the same stuff:


m = dy/dx = - 2w
at x0 = w,  y0 = 2 - w^2.

So  

y - (2 - w^2) = -2w(x - w)

y - 2 + w^2 = -2wx + 2w^2

y = 2 - 2wx + w^2


Now the intercepts are:

y-int:  let x = 0,  y = 2 + w^2  --- your height
x-int:  let y = 0,  solve for x:

0 = 2 - 2wx + w^2

2wx = 2 + w^2

     2 + w^2
x = ----------    --- your base.
       2w

                           2 + w^2
Your triangle has  A = 1/2 --------- (2 + w^2)
                              2w
   ( 2 + w^2)^2
A = -------------
        4w

Now minimize that.  I assume you have learned max-min problems, so you can finish up. You will need the quotient rule, of course.