Skip to content
Question 40 of 47

Q.(a) An economy produces only coal and steel. These two commodities serve as intermediate inputs in each other's production. 0.4 tonne of steel and 0.7 tonne of coal are needed to produce a tonne of steel. Similarly 0.1 tonne of steel and 0.6 tonne of coal are required to produce a tonne of coal. No capital inputs are needed. Do you think that the system is viable? 2 and 5 labour days are required to produce a tonnes of coal and steel respectively. If economy needs 100 tonnes of coal and 50 tonnes of steel, calculate the gross output of the two commodities and the total labour days required.

(OR)
(b) Verify Euler's theorem for the function u=x3+y3+3xy2u=x^3+y^3+3xy^2.
Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2024Subjective· 5mImportance★★★★★
85% · 40/47 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) The economy is viable; gross outputs steel ≈176.47\approx176.47 t, coal ≈558.82\approx558.82 t, needing 20002000 labour days. (b) Euler's theorem holds: xux+yuy=3uxu_x+yu_y=3u.

(a) Input–output (Leontief) model

Order (steel, coal). 1 t steel needs 0.40.4 steel, 0.70.7 coal; 1 t coal needs 0.10.1 steel, 0.60.6 coal. Technology matrix (columns = producing sector):

A=[0.40.10.70.6].A=\begin{bmatrix}0.4&0.1\\0.7&0.6\end{bmatrix}.

Viability. I−A=[0.6−0.1−0.70.4], det⁡=(0.6)(0.4)−(−0.1)(−0.7)=0.24−0.07=0.17.I-A=\begin{bmatrix}0.6&-0.1\\-0.7&0.4\end{bmatrix},\ \det=(0.6)(0.4)-(-0.1)(-0.7)=0.24-0.07=0.17. Diagonals 0.6,0.4>00.6,0.4>0 and det⁡>0\det>0 → viable.

Gross output X=(I−A)−1D, D=[50100]X=(I-A)^{-1}D,\ D=\begin{bmatrix}50\\100\end{bmatrix} (steel 50, coal 100):

(I−A)−1=10.17[0.40.10.70.6],X=10.17[0.4(50)+0.1(100)0.7(50)+0.6(100)]=10.17[3095].(I-A)^{-1}=\frac1{0.17}\begin{bmatrix}0.4&0.1\\0.7&0.6\end{bmatrix},\quad X=\frac1{0.17}\begin{bmatrix}0.4(50)+0.1(100)\\0.7(50)+0.6(100)\end{bmatrix}=\frac1{0.17}\begin{bmatrix}30\\95\end{bmatrix}.

Xsteel=300017≈176.47 t,Xcoal=950017≈558.82 t.X_{\text{steel}}=\frac{3000}{17}\approx176.47\text{ t},\qquad X_{\text{coal}}=\frac{9500}{17}\approx558.82\text{ t}.

Total labour (coal 2 days/t, steel 5 days/t): …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.