36 kg bag of cement was dropped losing one tenth of cement. How much was left in the bag of cement ?
2 answers:
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Answer:
$15.625
Step-by-step explanation:
Let the revenue collected be R and price per spectator be p then the number of spectators be N. Therefore
R=Np
Using equation of slope of y=mx +c where m is gradient and c is y-intercept
When p=$11, N=27000 and when p=$8, N=31000
The gradient, m will be
To get the y-intercept
N=-1333.33p+c
When spectator number n is 27000, the price p is $11
27000=-1333.33(11)+c hence we solve c
c=27000+(11*1333.33)= 41666.67
Therefore, the linear equation is
N=-1333.33p+ 41666.67
Substituting the linear equation into R=Np we obtain
R=p(-1333.33p+41666.67)
To obtain maximum revenue, we differentiate the above with respect to price hence obtain
0=2*-1333.33p+41666.67
Therefore, the price that maximizes revenue is $15.625