I have two problems that I need explaining for

1. The height of a cylinder is four times its radius. Find a function that models the volume v in terms of the width w.

10. A rectangle's base is on the x-axis and its other two vertices are above the x-axis, lying of the parabola y = 8 - x^2. Find the base and height that give the largest area for the rectangle.

does the width mean its diameter?

I believe so

1) the volume of a cylinder formula is V = (pi)(d/2)^2 * h
you also know that h = 4r = 4(d/2) = 2d. So plug in 2d for h:
V = (pi) (d/2)^2 * (2d) = (pi)(d^(2)/4) * (2d) = (1/2)(pi)d^3

E: change the d's to w's

This post was edited by TritonV8 on Sep 29 2013 06:23pm

Thanks

if i understand the question right you are looking for the rectangle with the largest area fitting below the 'cone' described by the parabola equation, with the two top corners touching the line of the parabola (which is symmetric around the line x=0) - ie you are looking for the maximun of the function 2x(8-x^2)

This post was edited by brmv on Sep 29 2013 06:59pm

so do guess and check?

This post was edited by aman1 on Sep 29 2013 07:00pm

can you use wolfram?

postscript:

with 'can' i mean
1. are you allowed to
2. do you have access

btw, my best guess is that it is the square from x=-2 to x=+2 going up 4 units, ie 16 square units

This post was edited by brmv on Sep 29 2013 07:27pm

#10

Gameplan:
1. Express the area of the rectangle as a function of a single variable
2. Graph this function
3. By looking at the graph and plugging numbers into the function, estimate the value that maximizes the function (the area)
----------------------------------------------------------------------------------------
1. f(x) = 8-x^2
The vertex of this parabola is on the line x=0. So the base of the rectangle will be 2x. The height of the rectangle will be f(x)
So the area of the rectangle will be base * height, or:
f(x) * 2x
or (8-x^2)*2x
or 16x - 2x^3
so that is A(x) = 16x - 2x^3

3. If you graph that, you'll find it has 2 local maximums NEAR x = -1.6 and x = 1.6
So the base would be about 2*1.6, and the height would be about f(1.6)=5.44, and the maximum area would be about 17.4 (square units).

With calculus we could dispense with all these estimates and get the actual values, but I don't know a way to calculate the maximum exactly without it.