Thomaskutty is starting a new business. he knows that most business when they start run at loss. Thomaskutty predict that in the first month he will lose \$2200, in the second he will loose \$1800 and in the third he will lose $1400 and this pattern will continue with each passing month (like arithmetic sequence where a=-2200 and d=400). Given the cumulative (running) total profit after n months is given by S(n)=n/2 (2a+(n-1)d) use calculus to determine:
a) The last month Emma will incur a net loss
b) The amount of loss during that month (given tn=a+(n-1)d)
c) The maximum cumulative loss (minimum cumulative profit) that Emma will experience
My working for this is quite stupid in my opinion. I try to work out these question by using the arithmetic sequence formula when we are asked to use calculus. I am not sure where to start to use calculus to solve this Here is my working out:
a) $tn+(n-1)d$ = -2200 + (7-1)*400 = 200
Thus during the seventh month it will be the start of the positive.
-2200 + (6-1)*400 = -200 In the sixth month it will be the last month of a loss.
b) the loss on the last month will be \$-200.
c)The maximum cumulative loss by Emma will be
n/2 (2a+(n-1)d) 6/2 (2(-2200)+(6-1)*400 = \$-7200 loss
Is my working out for the answer correct and if they are how can I use calculus to solve the questions?
Since we are using a calculus method, let's use the function $p(t)=-2400+400t$.
I start with $-2400$ because averaged over a month, adds up to only $-2200$ negative.
a. $p(t)=0$, which means $t=6$. So her last month of net loss is the sixth month (she stops losing money at $t \geq 6$, which is the start of the seventh month.
b. $\displaystyle \int_5^6p(t)\,dt=-200$, or a $\$200$ loss. You can reaffirm this with the arithmetic sequence
c. The maximum cumulative loss is at the point where she stops losing money further, or at $t=6$. This amount is $\displaystyle \int_0^6 p(t)\,dt=-7200$, or a $\$7200$ loss.