# Maximum Profit at which combination

A dealer deals only in colour TVs and VCRs . He wants to spend up to Rs.12 lakhs to buy 100 pieces. He can purchase colour TV at Rs. 10,000 and a VCR at Rs. 15,000. He can sell a colour TV at Rs. 12,000 and a VCR at Rs. 17,500. His objective is to maximiz profits

Fpr the maximum profit number of colour TVs and VCRs that he should stock is

I have found answer using hit and trial But is there a mathmatical way to guarantee that maximum profit happens for 60 colour TV and 40 VCR

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its a question which needs careful analysis –  Rajesh K Singh Aug 4 '12 at 14:37

Yes. You must consider the boundary conditions. If we let $t$ denote the number of TVA sold and $v$mdenote the number of vcrs sold, the we know that $t+v=100$, and $10000t+15000v \leq 1200000$.

Graphing these lines on the $t,v$ axes, along with the lines $t =0, v=0$, we get an enclosed region representing all the possible sales combinations. If you think about it, you can determine that the optimal point will always be on the boundary. And because all the relationships are linear, the optimal point will be at the intersection of two boundary conditions. (otherwise you shift along one of the lines in whichever direction increasesrofits)

Plugging in $t = 100 - v$ into $10t+15v=1200$, we get $1000 +5v = 200$, so that their intersection is at $(40,60)$. Comparing the profits from only selling TVA, only selling vcrs, selling nothing, and selling $(40,60)$, we see that it's the latter that has the best profit.

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assuming total number of pieces of CTVs and VCRs together is $c$ which is a constant once the purchase is made.

let, $x$ is the number of CTVs sold(purchased), and $(c-x)$ is the number of VCRs sold(purchased)

profit function, $f(x)=(x)(2)(10^3)+(c-x)(\frac{5}{2})(10^3)$ -------(1)

cost price, $g(x)=(x)(10)(10^3) + (c-x)(15)(10^3)\le(12)(10^5)$ -------(2)

integer, $c = 100$ -------(3)

maximum number of CTVs which can be purchased, $x = \frac{(12)(10^5)}{(10)(10^3)}=120$

maximum number of VCRs which can be purchased, $(c-x) = \frac{(12)(10^5)}{(15)(10^3)}$

i.e. $0 \le x\le 100$ -------(4)

we have, $0 \le (c-x)\le 80$ -------(5)

we always need to check the boundary conditions which means consider the following equation

$g(x_b)=(x_b)(10)(10^3) + (c-x_b)(15)(10^3)=(12)(10^5)$ -------(6)

also, $f(x_b)=(x_b)(2)(10^3)+(c-x_b)(\frac{5}{2})(10^3)$ -------(7)

from equation (6),

$x_b = (3c-240)=60$

$f(x)=(2x+\frac{5}{2}(c-x))(10^3)$

$f(x)=(\frac{5c}{2}-\frac{x}{2})(10^3)$ -------(8)

if $x$ is the number of CTVs sold, then on solving, (1) and (2) we have

$f(x)\le(c+120)(10^3)$ -------(9)

i.e. profit $f(x)\le(220)(10^3)$, where $x$ is the number of CTVs sold(purchased)

also, from (8) and (9) we have $x_m$ the maximum number of CTVs sold(purchased) from the following equation

$(\frac{5c}{2}-\frac{x_m}{2})(10^3)=(c+120)(10^3)$

i.e. $x_m=60$

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NOTE : $5 CTVs \equiv 4 VCRs$ in terms of profit per piece –  Rajesh K Singh Aug 4 '12 at 11:05
NOTE: 100 $CTVs$ $\equiv$ 80 $VCRs$ in terms of profit per piece –  Rajesh K Singh Aug 4 '12 at 11:07
NOTE: 3 $CTVs$ $\equiv$ 2 $VCRs$ in terms of cost per piece –  Rajesh K Singh Aug 4 '12 at 11:14