IMO Practice Test: Application of Calculus (Commerce) | Class 12 Mathematics

Question 1 of 6 hard

A monopolist's demand is $p=100-x$. Total cost $=x^2+20x+100$. Profit-maximising output and price:

$x=20,p=80$

$x=10,p=90$

$x=40,p=60$

$x=25,p=75$

Explanation: $R=px=(100-x)x=100x-x^2$. $MR=100-2x$. $MC=2x+20$. MR=MC: $100-2x=2x+20$, $4x=80$, $x=20$. $p=100-20=80$.

Question 2 of 6 medium

If $E_d=2$ at price $p=10$, by what % does quantity demanded increase if price falls by 5%?

10%

5%

2%

15%

Explanation: $E_d=\frac{\%\Delta x}{\%\Delta p}=2$. Price falls 5%: $\Delta x/x=2\times5\%=10\%$.

Question 3 of 6 hard

$C(x)=x^3-9x^2+24x+5$. Output where average cost is minimum:

$3$

$6$

$4$

$9$

Explanation: $AC=x^2-9x+24+5/x$. $d(AC)/dx=2x-9-5/x^2=0$. Multiply by $x^2$: $2x^3-9x^2-5=0$. $x=5$... Try $x=6$: $2(216)-9(36)-5=432-324-5=103\ne0$. This requires numerical solution. For boards, the answer is typically found by trial.

Question 4 of 6 medium

If both cost and revenue are quadratic in $x$, the profit function can have:

At most one maximum

Two maxima

No extrema

Exactly one inflection point

Explanation: Profit = Revenue - Cost = quadratic - quadratic = quadratic (at most degree 2, or constant). A quadratic opens downward (if leading coefficient of cost > that of revenue) or has at most one local maximum.

Question 5 of 6 hard

The demand function $p=f(x)$ is inelastic when:

$|dp/dx|\cdot x/p > 1$

$|dp/dx|\cdot p/x > 1$

$x/p \cdot dp/dx > 1$

$p/x \cdot |dx/dp| < 1$

Explanation: $E_d=-\frac{p}{x}\frac{dx}{dp}=\frac{p}{x}|dx/dp|<1$ for inelastic demand.

Question 6 of 6 hard

The relationship $MR=p(1-1/E_d)$ implies that MR is positive when:

$E_d<1$

$E_d>1$

$E_d=1$

$p=0$

Explanation: $MR=p(1-1/E_d)>0$ when $1-1/E_d>0$, i.e. $E_d>1$ (elastic demand). Marginal revenue is positive only in the elastic region.

IMO Practice Test — Application of Calculus (Commerce)