Question

A one-product company finds that its profit, P, in millions of dollars, is given by the following equation where a is the amount spent on advertising, in millions of dollars, and p is the price charged per item of the product, in dollars. P(a.p)= 2ap + 80p – 15p? - Hoa?p-120 Find the maximum value of P and the values of a and p at which it is attained. The maximum value of P is attained when a is $1 million and p is $ The maximum value of P is $ million

Answer 1

In the problem a And p are amounts so both should be greater than zero , because it obvious that atleast some amount is spent for both a and p !

explanation :

Solution : given Plab) = 20$+80°-15.p2-ta q-p-120 Pla, p) profit in terms o of a, p amount spent on Advertisement. + price charged peu item This problem can be treated as find Maximumum / minimum value of A Multivariable function, So here multip Multivariable function is Pla, p) of variables a, p. the How to find Maximum & So, or minimum value of a aqulti variable function ? For Function Plaip). I Steps are :- ead Find al & OP ap © Equate op & OP da Find points (a, b) solving above D = 2² P 2² p Equations to 22p7 CS stanned find CamScanner da² 3.8² saopf

9. olio partially differentiation wanita taking p as constant. da So, IF :- Dla, p) so & 2² p Lo then. da2 Maximum, pla, p) - Dla, p) >0 & gap so then. 292 Minimum. pla, p) Dla, f) to pla, p) → saddle point, to So, Follow these step From P (a, b) = 2ap+80p - 15 p2 - I ape- ОР 2aß For 10 & de 29 480 - 30 ap 30 p - CS scannect with both equale to CamScanner

Day ep-zap & Jos 2a +80-30p-a² To So From D P[2-a 5 2p - ap p=0, a = 10 Sean From 2a +80-30p - a² put p=0 when p=0 10 Q: 2 - 2 at 80 92 2oat 800 - 92=0 a? 20a - 800 = ) a ² - 4 0 a +200- 800 = o. a la-40) +20 (a-40) = 0 (a+2o) (a-40) Scanned_with CS. CamScanne F -20, a = 40

20 & 40 So - a= For p=0. =) a. p) = (40, o o) - 20,0 Now For pr 74 a =10 [ put put a=po 29 +80-30 p - 9² 2x10 +8o-308- 10² 20 + 8.0- 30ß -10 3 go go=80p when a=10 (10,30 37 So, [a, po So, we have three critical point (9, ) = (-20,0) (40,0), (10,30 But shoud always 7o. a. p both are Amount ên dollars So, both should be greater than zero. Amount As cs Scanned with CamScanner

We coin So, point. Consider only one critical (0,5) = (10,5) D. E g-p.022 2°РТ aps Laaap 2 [2p-2ap 2 Now Find da² dups a op [2P da to 89² da da to te te I op. 2/29 +80-300 - 92 ӘР te ede ap de -30 22 p 2- 2a gep apda sebe P.-30 [ =) D (a, b) = CS 68 -[2-9 5 Scanned with CamScanner

check Marima & So put (a, b) to Conditim Minima 2 =) D (1,3) 107 18 2-2]" D (103) 18-0 =18. 'so D (10,3)=18 > a And Also, At PL 10,3) P. Hih 29² 3/s 2² p da² So, P410,3 is Maximum - Maximum value of Þ when a= p=1 3 7 - Maximum value of p will be p/103) 2x10x3 +8048-15x3² – 1 x Go +240 - 135 - 30-120 CS Scanned with Cams Maximum of P = 15 Ans Tom 10°33-120 - 15

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