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

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. $x_m=60$
