maximizing and related rates: rising water level; max area

These problems are done to death. You can google and find many examples. Maybe not your exact dimensions, but the same type of problem. You can also search this site and find past postings of the very cliche 'inverted cone' problem.
 
cosh said:
1) A water tank has the shape of an inverted circular cone with base radius 2 meters and height 4 meters. If water is being pumped into the tank at a rate of 2 m^3/min, find the rate at which the water level is rising when the water is 3 meters deep.
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Draw the side view of the tank, being an upside-down isosceles triangle. Draw the center-line from the top to the angle at the bottom of your drawing. Label the height and the radius (that is, half of the triangle "base") with the listed values. Draw another horizontal line, somewhere in the middle of the height, for the water level.

Label the water height and the water radius with variables. Using similar triangles (the whole-tank half-triangle and the water-level half-triangle), you can express the water-level radius in terms of the water-level height. Use this to create a function for the volume V of water at any given height.

You are given that dV/dt = 2, and you're asked to find the rate of change in the height when the height is 3. Differentiate your volume function, plug in the known values, and solve for dh/dt.

cosh said:
2) A rectangle has one side on the x-axis and two corners on the top half of the circle of radius 1 centered at the origin. Find the maximum area of such a rectangle. What are the coordinates of its vertices?
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Draw the half-circle and the rectangle. Note that the rectangle with maximal area will be twice as large as the maximal rectangle with one side on the y-axis and being only in the first quadrant. (Use this to simplify the calculations.)

You know that the width of this (half-size) rectangle will be "x". You know the equation of the half-circle, and this gives you the y-value for any x-value. In particular, this gives you the expression for the height of the rectangle for any given x-value.

Use this information to create an area function for the rectangle. Then maximize the area.

If you get stuck, please reply showing (or describing) all of your work and reasoning. Thank you! :D

Eliz.
 
The equation for a circle of radius 1 is \(\displaystyle \L\\y=\sqrt{1-x^{2}}\)

The height of the rectangle is y and length 2x

The area of the rectangle is A=2xy.

Now, you have \(\displaystyle \L\\A=2x\sqrt{1-x^{2}}\)

Differentiate, set to 0 and solve for x.

circlebc6.gif
 
cosh said:
I'm trying to understand why the length would be 2x.
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Look at the drawing and re-read the explanation. What is the length of the half of the rectangle that's in the first quadrant? So what is the length of the whole rectangle? :wink:

Eliz.
 
As galactus said, you set the derivative equal to zero to find the maximum area of the rectangle - given by the intercept of the graph of the derivative.
\(\displaystyle \L \frac{dA}{dx} = 0 = 2\sqrt{1-x^{2}} - \frac{2x^{2}}{\sqrt{1-x^{2}}\)

Because of the zero, you won't have to do too much tedious algebra to solve for x as you can just multiply both sides by \(\displaystyle \sqrt{1-x^{2}}\) so you won't have to deal with the fractions and the square roots.
 
You forgot to multiply the \(\displaystyle 2\sqrt{1-x^{2}}\) with \(\displaystyle \sqrt{1-x^{2}}\). That would've gotten rid of all square roots and fractions ... well, it DOES get rid of them :wink:
 
You have an equation you have an x value for maximum area (has to be positive) and an equation \(\displaystyle A = 2x\sqrt{1-x^{2}}\). Oh what oh what can we do? Hmmmm.

(I'm assuming you're referring to the coordinates of the absolute/local maximum of the graph).
 
Oh sorry. I didn't see the last line of your question. Yes, you plug in the value you got for x into \(\displaystyle y = \sqrt{1- x^{2}}\) to find your y-value and yes, your two upper coordinates will be (x,y) and (-x,y). Don't forget the vertices on the x-axis!
 
Look at galactus's diagram. It should be pretty clear what the y-coordinate should be and you already know what the x-coordinates are.
 
(-1,0) and (1,0) are points on the circle! Not points of the rectangle.

Rectangles have 4 vertices. For this question, two of the vertices are made so that they have to touch the circle while the other two must stay on the x-axis. For the purposes of finding what the other vertices are, imagine that the circle just wasn't there anymore.

What are the two points on the x-axis that make up the rectangle? You know what the x-coordinates are and you know that these two particular vertices LIE on the x-axis (what does this tell you about the y-coordinate if it LIES on the x-axis?).
 
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