Question

please answer areas with red x. i have tried these value positive and negative and they are not correct.

GreenLawns provides a lawn fertilizing and weed control service. The company is adding a special aeration treatment as a low-cost extra service option, which it hopes will help attract new customers. Management is planning to promote this new service in two media: radio and direct-mail advertising. A media budget of $4,000 is available for this promotional campaign. Based on past experience in promoting its other services, GreenLawns has obtained the following estimate of the relationship between sales and the amount spent on promotion in these two media: S = -2R2 – 12M2 - 9RM + 18R + 33M, Where S = total sales in thousands of dollars R = thousands of dollars spent on radio advertising M = thousands of dollars spent on direct-mail advertising GreenLawns would like to develop a promotional strategy that will lead to maximum sales subject to the restriction provided by the media budget. (a) What is the value of sales if $2,000 is spent on radio advertising and $2,000 is spent on direct-mail advertising? $ 91886000 (b) Formulate an optimization problem that can be solved to maximize sales subject to the media budget of spending no more than $4,000 on total advertising. Max -2 p2+ -12 M2+ JRM -9 + 20 37 s.t 1 + g 1 s 4 R, M 2 0 (c) Determine the optimal amount to spend on radio and direct-mail advertising. How much in sales will be generated? Amount spent on radio advertising = $ 4000 Amount spent on direct-mail advertising = $ 0 Total Sales = $ 11953985.5

Answer 1

Note: All the parts have been done as shown in the attached images. In case of any doubt, please mention it in the comment. Thank you.

Page where Be Themen hemen advertising Sales -22_19M8-9RM +18R+33M dollars spent on radio adrest M- Thousands dollars pent direct mail advertising of spent Amount spent If Given amount on radio advertising =>$bodo on direct mail advertising → $ 2000 Therefore, R= 2 and M=2 Sales= -2X(2)*12/23_9x2x2+18x2+ 33x2 = $ 10 (in thousands) Sales = $ 10,000 CB Max. 7 R & ARRO (6) Max. -2R2 2 _ 12M2_9 RM + 18R+ 33M AR+ M L 4 RM zo Scanned by TapScanner

Date Page co We can Solve this using lagrange multipliers fCRM)= -28_12M298M +18R +33M R+M - 4 g (RM) Iccording to Jag range multipliers dll dR s do CM) =0 dR - s dg and def(RM) – (RM)=0 dM Therefore df CRM) = -48–9M +18 CM) - y IM هام -9R+33 1 dg (B, M) d deckM) -1 dM Scanned by TapScanner

Date Page So -4R-9M + 18-1XX=0 and -24M - 9R +33 - 1x) - 0 -4R-9M +18 = -20-9R +33= from 7 & 1 -4R-9M + 18 = -20 9R + 33 5R +15M-15 = 0 R+3M-3=0 TTT. But we have R+M = 4 Put value in (1) of R from 4-M +3M-3 =0 4 1+2M=0 M =-0.5 Since value M ca of can not be therefore negative, Scanned by TapScanner

Date Put value of Mi Put value of M in TV D +R=4 of R=4 * Put value R & M in salu function is Sales = -2XC4) 12x 69_9x480 +18X4 +33X = 40 Cin thousands) Therefore, co Radio R: $ 4000 Direct Mail M=$0 Total Sales => $ 40.000 Scanned by TapScanner