1. ## speed and rate

A small office consumes electricity at a rate represented by the equation y = 36000t – 3600t2 where y is the rate of consumption of electricity in watts (W) and t is the time in hours. Assuming that the office runs for 10 hours a day, find the cost of electricity bills of the office in 1 day. You may assume that electricity is charged at S$0.005/W. 2. Originally Posted by dorwei92 A small office consumes electricity at a rate represented by the equation y = 36000t – 3600t2 where y is the rate of consumption of electricity in watts (W) and t is the time in hours. Assuming that the office runs for 10 hours a day, find the cost of electricity bills of the office in 1 day. You may assume that electricity is charged at S$0.005/W.
Y is the rate of consumption of electricity in (Watt) so and t in hours
the office runs for 10 hours a day ..

in one day 10 hours runs so just substitute 10 in the equation you will find how many watt consumption in one day then just multiply the watt consumption in one day with the cost of the watt S\$0.005/W you will have the answer

$y=36000t-3600t^2$

you can solve it it is easy try

3. but when i substitute t = 10 into the equation,
i get 0 watt. so how can i find out?
i really have no idea.

4. the question is not logic because the consumption of electricity increasing until it reach 5 hours then it begin to decreasing until it reach 10 where the consumption of electricity equal zero ...or

I think I misunderstand the question because he said the rate of consumption of electricity so if we need the consumption of electricity we should integrate

$\int 36000t-3600t^2 dt=18000t^2-1200t^3 +c$

but c=0 because if the office dose not run then the consumption of electricity would be zero finally you have

$\text{consumption of electricity}=18000t^2-1200t^3$

sub 10 in the equation

$18000(100)-1200(1000)=600000$

the cost

$600000\times\frac{5}{1000}=3000$

I am sooooo sorry I like to help others but I missed sometimes I should stop solving questions and leave them for whom are better than me ......

Amer

5. you shouldn't said that. you are really generous enough to put in effort to help others though u do not know them.
and your answer is correct as i've checked against the ans sheet.
• but could you explain why the consumption of electricity increased until it reach 5 hours and den decrease when it reach 10hrs?
• why do we have to integrate to find consumption of electricity?
• and why c = 0?

6. Ok

the question said the rate of change like the acceleration do you know it the acceleration you can find the rate of it

$A=\frac{v_2-v_1}{t_2-t_1}$

forget it the question as I said give you the rate of consumption that's mean he gave you how the consumption change with the time
in the first hour the rate of consumption =36000-3600
and when the rate is zero that mean there is no change in the consumption in that hour but not mean there is no consumption

about what I said that the rate of consumption increase until it reach 5 then it began to decrease if you find the solution (where it is zero) for the derive rate of consumption then find where it is positive and where it is negative the derive of rate of consumption can given by

36000-3600(2)t find the solution of it you will find the solution is 5 then take numbers bigger than 5 and sub it in (36000-3600(2)t) if the value is positive you put that mean that the rate of consumption is increasing in the interval (5,infinity) take another value less than 5 sub it if it is negative that mean that the rate of consumption decreasing that is it

c is zero in the

$18000t^2-1200t^3+c$
this formula give the consumption after hours example after how many hours if it runs 2 hours sub 2 in the equation then it give you how watt it consumption ok after zero hour how many watt the office consumption it is zero right then sub zero in the equation above the answer should be zero so c equal zero that is all I think it is clear now right ??

best wishes .......