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Differential of an Equation

Date: 01/02/2003 at 10:39:35
From: Dawn
Subject: Differential of an equation

How do I calculate the differential of an equation, to be used in 
finding the minimum value of the curve the equation describes? 

The curve describes the total cost of an inventory (Y) when orders are 
placed at X Units per order. Other elements in the calculation are 
constants: d, the annual demand for the item, in units; o, the cost of 
placing a single order, in dollars per Order; p, the price per Unit; 
and r, the annual cost of carrying a Unit of Inventory, expressed as a 
percentage of its value (for practical purposes, the interest, taxes, 
and insurance on the item).  With these elements, the formula for 
total cost is given by this equation:

   Y =(2od + rpx^2)/2x    (where x^2 means x squared)
If it helps, this is the sum of the two costs of an Inventory - the 
cost of ordering, and the cost of carrying it. They can be expressed 
separately as:

   Cost of a Year's Orders =  o * (d/x) =  od/x
     (This takes the annual demand and divides it by the 
      Order Quantity, to calculate the number of orders, then 	
	  multiplies by the Cost per Order.)

   Cost of carrying Inventory = rp * (x/2) = rpx/2	
     (This assumes that Inventory during the year goes 
      from 0 to the Order quantity, then gets used up evenly
      over time, and is replenished exactly when it gets 
      down to 0 again, so that the average Inventory on hand
      during the year is half an Order. Then it multiplies 
      that by the price per Unit, and the Cost of Carry per Unit.)

So now to find the value of X where Y is at a minimum, we need to 
differentiate the function above. According to the boo, the minimum 
value is given by the formula

   x = Square Root of the Quotient of (2od divided by rp)  or  

I can't get there from here.  Am I missing something?  

Date: 01/02/2003 at 22:59:00
From: Doctor Jeremiah
Subject: Re: Differential of an equation

Hi Dawn,

The differential operator is what is used to take the differential of 
a function, similar to the way a square root operator takes the square 
root of its contents. The differential operator looks like the d() 
function where the contents are to be differentiated.

So the differential of Y = (2od + rpx^2)/2x can be taken by applying 
the differential operator to both sides of the equation:

   d( Y ) = d( (2od + rpx^2)/2x )

Unfortunately this is nasty looking, so we should try to simplify the 
equation as much as possible. That means distributing the denominator 
so that we end up with 2 terms:

   d( Y ) = d( od/x + rpx/2 )

Much better. The differential of two objects added to each other is 
the sum of the two differential results, so:

   d( Y ) = d( od/x ) + d( rpx/2 )

The differential of a variable multiplied by a constant is the 
constant multiplied to the differential result of the variable, so:

   d( Y ) = od d( 1/x ) + rp/2 d( x )

Now the interesting bit. The differential of Y is dY and the 
differential of x is dx, but what about the differential of 1/x? The 
easiest way to deal with 1/x is to change it to an exponent (which 
would make it x^-1):

   d( Y ) = od d( x^-1 ) + rp/2 d( x )

The differential of a variable to a constant exponent is the value of 
the exponent multiplied by the variable to a one smaller exponent 
multiplied to the differential result of the variable base.  
d( x^a ) = a x^(a-1) d( x ) so:

   d( Y ) = od (-1) x^-2 d( x ) + rp/2 d( x )

The differential of Y is dY and the differential of x is dx so:

   dY = -od/x^2 dx + rp/2 dx

The slope of the graph is the change in the dependant variable (Y in 
this case) divided by the change in the independent variable (x in 
this case). The differential of some variable is an infinitely small 
change in that variable. So if we divide both sides by dx to get the 
slope of the graph, then:

   dY/dx = -od/x^2 + rp/2

Check out the difference between the equation and its slope:

       Y =  od/x   + rpx/2
   dY/dx = -od/x^2 + rp/2

Basically an x in the numerator becomes an x with a one smaller
exponent and one in the denominator becomes an x with a one larger 

Anyway, to find the minimum of the slope, you have to imagine what a 
graph looks like. The lowest value has higher values on both sides.  
So the slope on the left is downward (negative) and the slope on the 
right is upward (positive). Now, for the slope to change from negative 
to positive it must go through zero. The lowest point is located where 
the slope is zero. So to find the lowest point we must set the slope 
to zero:

                0 = -od/x^2 + rp/2

And then we solve this for x:

                0 = -od/x^2 + rp/2
           od/x^2 = rp/2
          2od/x^2 = rp
              2od = rp x^2
           2od/rp = x^2
   sqrt( 2od/rp ) = sqrt( x^2 )
        minumum x = sqrt( 2od/rp )

Now x is the value at the lowest point. To find the value of the 
inventory at that point we must substitute that into the inventory 

   Y = od/x + rpx/2   <===   x = sqrt( 2od/rp )
   Y = od/sqrt( 2od/rp ) + rp sqrt( 2od/rp ) /2

Which is a big mess.  We need to simplify this a bit:

   minumum Y = sqrt( 2odrp )

I hope that wasn't too confusing. Please write back if there is
something you feel I could clarify more.

- Doctor Jeremiah, The Math Forum 
Associated Topics:
College Calculus

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