##### Optimization Of Business Models Help

 Calculus Tutor: None Selected Time limit: 1 Day

Apr 19th, 2015

a. We have a point (1400,480) and a slope (-27/270) or -1/10. p(x)=-1/10x+620.

b. Maximize R=PQ=Q(-1/10Q+620)=-1/10Q^2+620Q. This happens at vertex -b/2a or -620/(-2/10)=6200/2=3100. So Q=3100, thus P=310. 480-310=170 dollars. Thus the rebate should be 170 dollars.

c. Maximize Profit=R-C=-1/10Q^2+620Q-(112000+160Q). Dropping constant yields (-1/10Q^2+460Q). Vertex is -b/2a or -460/(-1/10)=4600 units. That is, P=160. 480-160=320 dollars of rebate.

Apr 19th, 2015

the answer to question c is incorrect but the rest are all good

Apr 19th, 2015

I guess I was in a rush. Maximize profit=R-C=-1.10Q^2+620)=(112000+160Q). Dropping constant yields (-1/10Q^2+460Q). Vertex is -b/2a=2300 units. That is, P=390. 480-390= 90 dollars of rebates.

Apr 19th, 2015

thank you!

Apr 19th, 2015

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Apr 19th, 2015
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Apr 19th, 2015
Dec 6th, 2016
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