Question

The marketing manager of a department store has determined that revenue, in dollars, is related to the number of units of television advertising, x, and the number of units of newspaper advertising, y, by the function R(x,y)=850(87x−2y^2+3xy−3x^2). Each unit of television advertising costs $1000, and each unit of newspaper advertising costs $500. If the amount spent on advertising is $11500, find the maximum revenue.

Answer 1

The marketing manager of a department store has determined that revenue, in dollars, is related to the number of units of television advertising, x, and the number of units of newspaper advertising, y, by the function R(x,y)=850(87x−2y^2+3xy−3x^2). Each unit of television advertising costs $1000, and each unit of newspaper advertising costs $500. If the amount spent on advertising is $11500, find the maximum revenue.

solutions Given Roy) 850 (878-9y2 + 3xy –3x2) No. of units of television Advertising go no. of units of newspapers Advertising Each unit of television Advertising Costs $1000 Each unit of newspaper Advertising costs $500 Amount spent on Advertising is $11500 1000x + sooy = 11500 need to and the maximum reverse for critical points of Realy) = Rx = 0 &e Ry zo Roxe = 850 (87 +3y-Gs) Ry = 85064+3%) Rox = 850 (6) Ryy = 850(-4) Ray = 850(3) Ee Ry 850 (87+ 3y -6x)=0 850thy + 3) =0 8734-6 -Hy + 3x = 0 Now -620 3x=44 G2 = 84 substitute o o 8++ту - 89 го 87 = 87 =) y = 5y s

Now 6 84 6x = 8y =) <= 8(8715) =) x i) 6 116 -) ca 5 0=(Rxa)(Ryy) -(Rxy)! (6*850) (-4*850) - (850*3)² D = 6 = 850?(24-9) - 8502*15= 10837500 >0 Doo Ee E Rxx = -6 * 850 20 116 87 ) there exist a maximum S 5 reveuse R 87 5 =857820 a. Maximum Revenue is $857 1820 Like please