An appliance store manager estimates that for x television ads run per day, $R ( x ) = - 0.01 x ^ { 3 } + x ^ { 2 } - 3 x + 200$ refrigerators will be sold per month. Find $R ^ { \prime } ( 4 )$ and interpret what it tells us about sales.
A) $R ^ { \prime } ( 4 ) = 4.52$ sales will be increasing at about 5 refrigerators per month per ad when they're running 4 ads.
B) $R ^ { \prime } ( 4 ) = 203.36$ the cost of refrigerators will be rising by $203.36 if they're selling 4 per day.
C) $R ^ { \prime } ( 4 ) = 4.52 \text {; }$ they'll sell about 5 refrigerators if they run 4 ads per day.
D) $R ^ { \prime } ( 4 ) = 203.36 \text {; }$ they'll sell about 203 refrigerators if they run 4 ads per day.

Question

Quizplus · Accepted Answer

To find $R'(x)$, we differentiate $R(x) = -0.01x^3 + x^2 - 3x + 200$ to get $R'(x) = -0.03x^2 + 2x - 3$. Substituting $x = 4$ into $R'(4)$ gives $R'(4) = -0.03(4)^2 + 2(4) - 3 = -0.48 + 8 - 3 = 4.52$. This indicates the rate of change of refrigerator sales with respect to the number of ads, meaning sales are increasing at a rate of about 4.52 refrigerators per month per additional ad when 4 ads are run.

An Appliance Store Manager Estimates That for X Television Ads R(x)=−0.01x3+x2−3x+200R ( x ) = - 0.01 x ^ { 3 } + x ^ { 2 } - 3 x + 200R(x)=−0.01x3+x2−3x+200

An Appliance Store Manager Estimates That for X Television Ads $R ( x ) = - 0.01 x ^ { 3 } + x ^ { 2 } - 3 x + 200$