Subdividing a Field; Falling Water Level

Date: 7/28/96 at 17:34:52
From: Sunnie
Subject: Calculus

I am having trouble with the following problems: 

(1) A man wishes to put a fence around a rectangular field and then 
subdivide this field into three smaller rectangular plots by placing 
two fences parallel to one of the sides. If he can only afford a 
thousand yards of fencing, what dimensions will give him the maximum 
area?

(2) A water tank has the shape of a right circular cone of altitude 
twelve feet and a base radius of 4 feet. The vertex is at the bottom 
of the tank. If water is being taken out of the tank at a rate of 10 
cubic feet per minute, how fast is the water level falling when the 
depth is 9.5 feet?
  
If you could lead me in the right direction with these problems, it 
would be most helpful. 

Thank you, Susie.

Date: 7/28/96 at 21:34:0
From: Doctor Robert
Subject: Re: Calculus

(1) Imagine a rectangle with length y and width x.  Then the total 
amount of fence used in the corral you describe is 2y+4x (counting 
those dividers!)  So, 2y + 4x = 1000.  The area of the rectangle is 
A = xy.

Solving the first equation for y and substituting into the second we 
get  A = 500x - 2x^2.  

You want to maximize A, so take the derivative and set it equal to 
zero in order to get the value of x which maximizes the area.  Once 
you get x, you should be able to figure what y is.  

(2) Now the cone. Let y be the distance from the vertex up to the 
water level. Let x be the radius of the cone at the water level. Now, 
the volume of a cone is 1/3 the area of the base times the height, so 
that in our terms, the volume of the water in this cone would be

   V = (1/3)pi*x^2*y.

Looking at the cone one can see that the following proportion holds:

   x/y = 4/12 so that x = (1/3)y

Substituting this into the equation for V we have

   V = (1/27)pi y^3.

Taking a derivative we get 

   dV/dt = (1/27)pi*3y^2*dy/dt.

Since dV/dt is 10 and y = 9.5, you should be able to solve for dy/dt 
which is the rate at which the height of the water is changing.

Good luck.

-Doctor Robert,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/