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Calculus BC: Applications of the Derivative

Problems

Optimization

Related Rates

Problem : Suppose Andrea decides to open a lemonade stand and her grandfather agrees to buy her all the supplies she needs. Now she has to decide what price she will charge for the lemonade. The more she charges per cup, the fewer cups people will buy. Andrea (who happens to be a mathematician) figures that the number of cups she will sell if she charges x cents per cup is approximately given by the function

c(x) = 1000e -x/5    

What should Andrea charge to maximize her profits? How much will she make?

Andrea's total profit is the number of cups sold times the number of cents received for each cup. This profit is given by the function

p(x) = 1000xe -x/5    

with derivative


p'(x) = 1000 xe -x/5 + e -x/5  
  = 1 - 1000e -x/5  

Solving p'(x) = 0 for x , we get x = 5 , so Andrea should charge 5 cents per cup. She would then sell 1000/e 368 cups of lemonade, making $18.40 in profit.

Problem : Find the minimum value of f (x) = - 2x 3 +3x 2 + 12x - 1 in the interval [1, 4] .

We first look for critical points.

f'(x) = - 6x 2 + 6x + 12 = - 6(x + 1)(x - 2)    

Thus f has only one critical point in the interval [1, 4] , at x = 2 . Evaluating f at the critical point and the boundary points, we find


f (1) = 12  
f (2) = 19  
f (4) = -31  

so f clearly takes on a minimum value on the interval [1, 4] of -31 at x = 4 .

Problem : Does the function g(x) = 3 - 2x attain a minimum value on the interval (1, 3) ?

The fact that the boundary points are not included in the interval is crucial here. If x = 3 was included in the interval, g would attain a minimum on the interval of -3 at x = 3 . As is, g does not attain a minimum on the interval, because given any x in (1, 3) , we always have

g < g(x)    

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