Friends Matt and Gary are planning to study together. Matt buys some pencils and erasers. Gary spends the same amount of money to buy thrice as many pencils and five less erasers. Now they are set for the group study. If cost of one eraser is two more than the cost of one pencil, what is the minimum number of pencils bought by Matt and Gary together?
I took the variables as -
$p$ (number of pencils bought by Matt)
$e$ (number of erasers bought by Matt)
then,
$3p$ (number of pencils bought by Gary)
$e-5$ (number of erasers bought by Gary)
also,
$C_p$ (Cost of a pencil)
$C_e$ (cost of an eraser)
Since they both spent the same amount of money,
$$ pC_p + eC_e = (3p)C_p + (e-5)C_e $$
and $$C_e = 2 + C_p$$
On solving we get,
$$\frac{5(2+C_p)}{2C_p} = p$$
The value that we need is $p+3p$. Therefore, we need the minimum value of $4p$ (Number of pencils bought by Matt and Gary)
$$\frac{10(2+C_p)}{C_p} = p$$
I then differentiated it with respect to $C_p$ and equated it to $0$, to find the minimum value. But after differentiation I got
$$\frac{-20}{C_p^2}$$
Which I can’t equate to $0$
Can you point out what’s wrong in this method, and suggest me a better solution, explaining the concept used in detail?
The answer given is $12$
From your equation:$$2pC_p=5(2+C_p)$$ calculate $C_p$: $$(2p-5)C_p=10$$ You know $C_p>0$, so $(2p-5)>0$. The minimum integer that satisfies that is $p=3$. Then the total number of pencils is $4p=12$.