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Super Toys Ltd. (STL) manufactures and sells toys. "Super car" is one of their popular models. The marketing department has estimated the demand function for the model to be linear. If the price was fixed at Frw 570 , the daily sales of the model would be 400 toys, whereas if the price was increased to Frw 820 , the daily sales would drop to 200 toys.

Data from the production department indicate that the incremental cost of producing q toys of the model is given by the equation;
$\Delta \mathrm{C}(\mathrm{q})=2 \mathrm{q}-570$ and that the daily fixed cost is Frw 1,100

(i) The revenue functions if q toys are sold.
(ii) The total cost function.
(iii) The daily break-even number of toys
(iv) The point elasticity of demand when the demand is 110 toys. Interpret the economic meaning of your result.

in Mathematics by Wooden (2,352 points) | 11 views

1 Answer

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Best answer
(i) Demand slope $=\underline{570-820}=-1.25$

$400-200$ Equation of demand

$P=-1.25(q-400)+570=-1.25 q+1070$
Revenue, $R=(1070-1.25 q) q=1070 q-1.25 q^{2}$

(ii) Total cost, $\mathrm{TC}=\int(2 \mathrm{q}-570) \mathrm{dq}$
$=q^{2}-570 q+C$
$C=$ fixed cost $=1,100$
$T C=q^{2}-570 q+1,100$

(iii) Profit, $\pi=1070 q-1.25 q^{2}-q^{2}+570 q-1100$
$=-2.25 q^{2}+1640 q-1100$
At B.E.P, profit $=0 \Rightarrow \Rightarrow-2.25 q^{2}+1640 q-1100=0$
$q=-1640 \pm \sqrt{1640^{2}}-4(-2.25)(-1100)$
$q=0.67$ or $q=728$


(iv) $P=1070-1.25 q$
$\underline{\mathrm{d} p}=-1.25 \mathrm{pdq}=\frac{1}{\mathrm{dp}}=-0.8$
$\mathrm{dq} \quad$ When $\mathrm{q}=110, \mathrm{p}=932.5$
Point of elasticity, $\mathrm{E}=\mathrm{p} \times \underline{\mathrm{d}} \underline{\mathrm{p}}$
$\begin{array}{ll}q & d p \\ = & \quad \underline{932.5} \times-0.8\end{array}$

by Wooden (2,352 points)

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