Concavity: The rising popularity of notebook computers is

[Jade]

There is a word problem with two parts.

The problem states: This rising popularity of notebook computers is fueling the sales of the mobile PC processors. In a study concluded in 2003, the sales of these chips (in billions of dollars) was projected to be:

. . .S(t) = 6.8(t + 1.03)^0.49

...for 0 < t < 4, where t is measured in years, and with "t = 0" corresponding to 2003.

The first question: Is the sale always increasing for 0 < t < 4?

My answer: I plugged in 0, 1, 2, 3, and 4 into the function to come up with YES, the sale is always increasing.

The second question: Show that, on the interval (0, 4), S is concave down.

I believe you need to find the derivative of the function to find concavity. I believe the derivative to be:

. . .S'(t) = 3.332(t + 1.03)^-0.51(1)

...according to the Chain Rule.

 

[tkhunny]

Re: Concavity

The first question is the sale always increasing for 0<=t<=4? My answer: I plugged in 0,1,2,3,4 into the function to come up with YES, the sale is always increasing.

Very bad. You have shown nothing in between those values. You must prove the ENTIRE interval, not just a few points. If you are using the word "concavity", you MUST have been introcued to the first derivative. This is what you need. Then you can observe the sign of the first derivative on the desired interval.

In this case:

S(t)=6.8(t+1.03)^0.49
S`(t) = 0.49*6.8*(t+1.03)^(-0.51) = \(\displaystyle \frac{3.332}{(t+1.03)^{0.51}}\)

Now, just ask yourself, is that EVER negative of zero on [0,4]. Don't forget to include the zero. That's not increasing.

 

[Jade]

Second Derivative

Part A is not increasing at zero

For Part B you would need to find the second derivative? Then set to zero to find possible point of inflection?

S''(t)=-1.69932(t+1.03)^1.51(1)

 

[tkhunny]

\(\displaystyle \frac{0.332}{1.03^{0.51}} > 0\)

That appears to be a place where S(t) is increasing.

Is that expression ever negative on [0,4]?

 

[Jade]

My intial response

was that it does not ever become negative. Am I missing something?