A feedlot operator has cattle that average 600 lb each and are gainiing bout 10 lbs a week price they could?

sold for this week is $.80 per pound, but is dropping by 1 cent per pound each week.





CHALLENGING PROBLEMS





a. show that the total price the operator could get for an average animal is a quadratic function of the number of weeks after the present and find particular equation.





c. how long should the operator wait to get the maximum amount per animal

A feedlot operator has cattle that average 600 lb each and are gainiing bout 10 lbs a week price they could?
We know the cows currently weigh 600 lbs.


The cows gain 10 lbs a week every week.


The price they could sell at now is 80 cents per pound and drops 1 cent per week.





Value of the cow:


weight x price





The weight of the cow is a function of time, we'll make t the time in weeks. t=0 is the first week.


weight = 600 + t*10


The price of the cow is a function of time as well as a function of weight. The price drops per week from a starting price of 80 cents, so:


price = 0.80 - t*0.01


The value of the cow is a function of weight and time. We can combine the above equations.


value = price x weight


value = (600 + 10t)(0.80-0.01t)


This is now your quadratic equation, its more obvious if you multiply it together.


value=480-6t+8t-0.1t^2


value=480+2t-0.1t^2


Thats your equation.





To get the maximum amount per animal, you want to find the peak in the equation above. This can be done by graphing or completing the square, or the quadratic equation.


I graphed it in my calculator and the answer is 10 weeks at a value of $490.


Lets try completing the quadratric solution.


value=-0.1t^2+2t+480


[-b +/- (b^2-4ac)^(1/2)]/2a


a=-0.1, b=2, c=480





Instead of actually completing the quadratic, we can find its vertex;


t=-b/(2a) = -2/(2*-.1) = 10 %26lt;- 10 weeks


Now substitute into the equation to find the value;


value=-0.1t^2+2t+480


value(10) = -0.1(10^2)+2*10+480 = 490 $

daisy