Help with Calculus (Optimization/Extreme Value Functions)?

Question

Let f(x) = 6 - (x^2). For 0 http://img.photobucket.com/albums/v512/spike_reznor90/diagram-1.jpg

a) Find A(1).
b) For what value of w is A(w) a minimum?

Thanks!!

Answer 1

Actually, 0
The point of tangency is (w , 6-w^2).
The slope at such a point of tangency is f'(w)= -2w.

The tangent line is y-(6-w^2) = -2w(x-w)
y = -2wx + w^2 + 6
....... the x intercept is (w^2 + 6)/2w.
....... the y intercept is w^2 + 6.
The triangle is formed by these intercepts.
Thus the area is half the product of these intercepts.
A(w) = (w^2 + 6)^2/4w.

a) The slope is -2. The point of tangency is (1,5).
The tangent line is y - 5 = -2(x - 1) ....... y = -2x + 7.
The y int. is 7 while the x int. is 7/2.
Thus the area is 49/4 = A(1) ... from the earlier formula.

§
b) A(w) = 1/(4w) *[w^4 + 12w^2 + 36]
... .. .. .. = 1/4 [w^3 + 12w + 36w^-1]
Then A'(w) = 1/4[3w^2 + 12 - 36w^(-2)]
= 3[w^4 + 4w^2 - 12]/4w^2
the zeros are: w^4 + 4w^2 - 12 = (w^2 - 2)(w^2 + 6)
Thus w = sqrt 2.

Answer 2

The slope of the tangent to f(x) = -2x
The equation of the tangent t o f(x) is y=-2x +b
6-w^2 = -2w +b so b = 6+2w-w^2
So y = -2x +6 +2w-w^2
When x= 0, y = 6+2w-w^2
When y=0, x = .5(-w^2+2w+6)
So A = (6+2w-w^2)(.5(-w^2+2w+6))/2
A(1) = (6+2-1)(.5(-1+2+6)/2 = 7*7/4 = 49/4 = 12 1/4
Find dAdw and set it = to zero. Find the root that lies between 0and 6. That value of w will produce the minimum valuue and it is approximately .689

Answer 3

a)
The slope of the tangent line is f ' (x) = -2x. So at w = 1, the line has a slope of -2 and passes through the point (1, 5). Write the equation for this line. The triangle has a height equal to this line's y-intercept, and a base equal to its x-intercept. Use these to find (1/2)base*height, the area.

b)
In general, for a point w the line is going to have a slope of -2w and pass through (w, 6 - w^2). Write the general equation for this line in terms of w. Find the general expression for the area in the same way. Use the derivative to find the minium.