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A newsboy purchases papers at $10$ cents and sells them at $15$ cents. However, he is not allowed to return unsold papers.

Let,

$m=$Paper purchased by newsboy

$n=10$

$m猢�10$

$p=$profit

$D=$demand of the day

So,

$E\left(P\right)=\underset{k=0}{\overset{10}{鈭�}}鈥�15\mathrm{Min}鈦�\{k,m\}P(D=k)鈭�10m$

$=15\underset{k=0}{\overset{m鈭�1}{鈭�}}鈥�kP(D=k)+15m\underset{k=m}{\overset{10}{鈭�}}鈥�P(D=k)鈭�10m$

Let $f\left(m\right)$be right hand side of above equation.

Therefore,

$f(m+1)鈭�f\left(m\right)=\left\{\begin{array}{l}15\underset{k=0}{\overset{m}{鈭�}}鈥�kP(D=k)鈭�15\underset{k=0}{\overset{m鈭�1}{鈭�}}鈥�kP(D=k)\\ +15(m+1)\underset{k=m+1}{\overset{10}{鈭�}}鈥�P(D=k)鈭�15m\underset{k=m}{\overset{10}{鈭�}}鈥�P(D=k)鈭�10\end{array}\right\}$

$=15\underset{k=m+1}{\overset{10}{鈭�}}鈥�P(D=k)鈭�10$

$=15\left(1鈭�\underset{k=0}{\overset{m}{鈭�}}鈥�P(D=k)\right)鈭�10$

$=5鈭�15\underset{k=0}{\overset{m}{鈭�}}鈥�P(D=k)$

Now we need to find the maximum value $m$

$F_D\left(m\right)=\underset{k=0}{\overset{m}{鈭�}}鈥�P(D=k)$

$=\underset{k=0}{\overset{m}{鈭�}}鈥�\left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10鈭�k}鈮�\frac{1}{3}$

If $m=2$

$F_D\left(2\right)=\underset{k=0}{\overset{2}{鈭�}}鈥�\left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10鈭�k}$

$=\left(\begin{array}{l}10\\ 0\end{array}\right){\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^{10鈭�0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^{10鈭�1}+\left(\begin{array}{l}10\\ 2\end{array}\right){\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^{10鈭�2}$

$=0.01735+0.08674+0.19513$

$=0.2992\left(<\frac{1}{3}\right)$

If $m=3$

$F_D\left(3\right)=\underset{k=0}{\overset{3}{鈭�}}鈥�\left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10鈭�k}$

$=\left\{\begin{array}{l}\left(\begin{array}{c}10\\ 0\end{array}\right){\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^{10鈭�0}+\left(\begin{array}{c}10\\ 1\end{array}\right){\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^{10鈭�1}+\left(\begin{array}{c}10\\ 2\end{array}\right){\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^{10鈭�2}\\ +\left(\begin{array}{c}10\\ 3\end{array}\right){\left(\frac{1}{3}\right)}^3{\left(\frac{2}{3}\right)}^{10鈭�3}\end{array}\right\}$

$=0.01735+0.08674+0.19513+0.26014$

$=0.5594\left(>\frac{1}{3}\right)$

Since $F_D$is an increasing function we concluded that $F_D\left(m\right)<\frac{1}{3}$for all $m鈮�2$and $F_D\left(m\right)>\frac{1}{3}$for all $m鈮�3$.

If, $f(m+1)-f\left(m\right)>0$for all $m=0,1,2$and $f(m+1)-f\left(m\right)<0$for all $m=3,鈥�.,9$

This means $f$gets its maximum when $m=3$

So, the newspaper boy should need$3$more to maximize the profit.

The newspaper boy should need $3$more to maximize the profit.