Question

Monthly sales of an SUV model are expected to increase at the rate of S′1t2 = -24t 1>3 SUVs per month, where t is time in months and S1t2 is the number of SUVs sold each month. The company plans to stop manufacturing this model when monthly sales reach 300 SUVs. If monthly sales now 1t = 02 are 1,200 SUVs, find S1t2. How long will the company continue to manufacture this model?

Answer 1

The company will continue to manufacture this model for 19 months.

The question is ill-formatted. The understandable format is given below.

An SUV model's monthly sales are anticipated to rise at a rate of

$S'(t)=-24^{1/3}$ SUVs each month, where "t" denotes the number of months and "S(t)" denotes the monthly sales of SUVs. When monthly sales of 300 SUVs are reached, the business intends to discontinue producing this model.

Find S if monthly sales of SUVs are 1,200 at time t=0 (t). How long will the business keep making this model?

Given $S'(t)=-24^{1/3}$ ... ... (1)

Condition (1) at t=0; s(t) =1200

We find S(t)=300 then t=?

a) We find S(t)

from $S'(t)=-24^{1/3}$

$\implies \frac{dS(t)}{dt}=-24t^{1/3} ~~[\because X'(t)=\frac{dX}{dt} \\\implies dS(t) = -24t^{1/3}dt$

On Integrating both sides

$S(t)=-18t^{4/3}+c ~...~...~(2)$

Now, at t=0 then S(t)=1200

So, from (2)

1200=0+c

⇒c=1200

$\therefore$from eq (2)

$S(t)=-18t^{4/3}+1200 ~...~...~(3)$

b) Considering that the firm intends to cease production of this model once monthly sales exceed 300 SUVs.

So, take S(t)=300

from eq (2) $300=-18t^{4/3}+1200$

$t^{4/3}=\frac{1200-300}{18}\\=\frac{900}{18}=50\\\implies t=50^{3/4}\\\implies t=18.8030$

Hence the company will continue 18.8020 months.

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