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