The revenue R (in dollars) generated by the sale of x units of a digital camera is given by (...

Question:

The revenue {eq}R {/eq} (in dollars) generated by the sale of {eq}x {/eq} units of a digital camera is given by {eq}( x-140)^2=-\frac {5}{7} (R-27,440) {/eq}. Approximate the number of sales that will maximize revenue.

a. Maximum revenue occurs at {eq}x = 27,440 {/eq} units.

b. Maximum revenue occurs at {eq}x= 3,920 {/eq} units

c. Maximum revenue occurs at {eq}x= 140 {/eq} units.

d. Maximum revenue occurs at {eq}x =392 {/eq} units.

e. Maximum revenue occurs at {eq}x=19,600 {/eq} units

Stationary Points and its description:

Let {eq}y=f(x) {/eq} be a function of one variable. Then {eq}x=a {/eq} is called a stationary point of the function if {eq}{\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}=0 {/eq} and that stationary point is said to be a point of minima if {eq}{\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right]_{x = a}} > 0 {/eq} and that critical point is said to be a point of maxima if {eq}{\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right]_{x = a}} < 0 {/eq}.

that critical point is said to be a point of in-flexion if {eq}{\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right]_{x = a}}=0 ,but {\left[ {\frac{{{d^3}y}}{{d{x^3}}}} \right]_{x = a}} \ne 0 {/eq}.

Answer and Explanation:

Here the given function is

{eq}\eqalign{ {\left( {x - 140} \right)^2} = - \frac{5}{7}\left( {R - 27440} \right) & \Rightarrow - \frac{7}{5}{\left( {x - 140} \right)^2} = R - 27440 \cr & \Rightarrow R = - \frac{7}{5}{\left( {x - 140} \right)^2} + 27440 \cr} {/eq}

Hence,

{eq}\eqalign{ \frac{{dR}}{{dx}} = 0 & \Rightarrow \frac{d}{{dx}}\left[ { - \frac{7}{5}{{\left( {x - 140} \right)}^2} + 27440} \right] = 0 \cr & \Rightarrow - \frac{{14}}{5}\left( {x - 140} \right) = 0 \cr & \Rightarrow x = 140 \cr} {/eq}

Hence the stationary point is {eq}\displaystyle x= 140 {/eq}

Hence: Maximum revenue occurs at {eq}x= 140 {/eq} units.

Option (C) is correct.


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