Question

1. Suppose we know that if a store wishes to stock and sell the quantity x thousands of a laser gaming mice product, the DEMAND PRICE for which it can sell them out is given by 50 P(x) = thousands of dollars per mouse. 0.01.x² +1 (i) Just look at the function! Does this price INCREASE or DECREASE as x gets larger? And does that make sense? (ii) Find the function R(x) which gives the rebenue obtained by selling out the x mice. And what are the units of R(x)? (iii) Next, find the function R'(x), the marginal revenue function. (iv) Last month, the store stocked and sold 2000 of the mice. So what was the Revenue? HINT: What was x? (iv) Now suppose the store plans to restock and sell 2001 mice next month. Assuming the demand function remains the same, find the additional revenue obtained by stocking that one more mouse. HINT: Again, what is the x-value now?

Answer 1

We have given the demand price as a function of x so we will use here general way to answer all the parts . In the first part we know as denominator increases the value decreases so it will decreases. In the next problem we need to know the value of R(x) and then we differentiate with respect to x and get the marginal value . And at the given demand x we calculated the R(x) value.

We have given that P(x) = 50 0.olx²+1 p(a) gets smaller. (1) A8 * gets larger the term in denominator (0.01x2 +1) gets larger so gets larger P(x) decreases. (i) In general we Rias x play So as a know that 80 R(a) = 50x 0.01x² + 1 (iii) We use the simple derivative rule a (W) = value udu de 80 12 dR (2) = R'(x) = (0.0122+1) d d (50%) - 50x da Tx (0.01x2 + 12 R!(Q) = 50 (0.0122 +1) - 50x (0.023) (0.01–2 +12 Ras= 0.5x2 +50-X 0.542-2 +50 (0.01x2 + 12 (0.01x2+12 R!(X) = 0.522-x+50 (0.01x2 +12 R( X = 2000) - 50 X 2000 2.49993 0.01X(2000)2 +1 R(x=2001) - 50 X 2001 - 2.498668 0.01x (2001)2 +1 so R(x=2001) - R(x=2000) = 0.001242