Bulk-Discount Revenue Kink at 200 Blades
Vidarbha Agro Tools sells a spare blade in lots. For orders up to 200 blades, revenue is R(x)=120x rupees. To encourage larger orders, the firm uses a bulk schedule: for x>200, each blade beyond 200 is priced at ₹100, and the reduction is modelled by R(x)=24000+100(x-200)-0.05(x-200)^2, where x is the number of blades in the lot. The accounts officer wants to know whether R is differentiable at x=200, because the marginal-revenue figure will be used in a budget forecast. A junior clerk has already computed R(200)=24000 from both formulas and concluded that differentiability is automatic. The production team is considering lots of 205 blades and wants an estimate of the extra revenue using differentials. There is no upper limit stated for x, but the quadratic term is intended only for modest bulk orders.
1. Test whether R is differentiable at x=200. Show continuity and compute both one-sided derivatives.
Read the worked answer
At x=200 the first branch gives R(200)=120×200=24,000. The second branch also gives 24,000+100(0)-0.05(0)^2=24,000, so R is continuous at 200. For x<200, R'(x)=120, so the left derivative at 200 is 120. For x>200, differentiate the second branch: R'(x)=100-0.1(x-200). Letting x→200+ gives 100. Since 120≠100, the one-sided derivatives differ. Therefore R is continuous but not differentiable at x=200; the graph has a kink there.
2. State the marginal revenue just below and just above 200 blades, and explain which value a forecast for a lot of 205 blades should use.
Read the worked answer
Just below 200 blades, marginal revenue is ₹120 per blade. Just above 200, it is ₹100 per blade. The abrupt drop occurs because the bulk price applies only to blades beyond 200. For a forecast moving from 200 to 205 blades, the relevant branch is x>200, so the right-hand marginal value ₹100 should be used. The left-hand value ₹120 would describe an increase from below 200 up to 200, not an increase beyond 200. At exactly x=200 there is no single derivative, so the phrase marginal revenue at 200 must be qualified by direction.
3. Use a differential based on the right-hand branch to estimate the revenue increase from 200 to 205 blades, and compare it with the exact increase.
Read the worked answer
Using the right branch, R'(200+)=100. For dx=5, the differential estimate is dR≈R'(200+)dx=100×5=₹500. Exact increase: R(205)-R(200)=[24,000+100(5)-0.05(25)]-24,000=500-1.25=₹498.75. The differential overestimates by ₹1.25. This is expected because the second branch has R''(x)=-0.1<0, so the revenue curve is concave; the tangent lies above the curve. The estimate is close because dx is small, but the exact calculation should be used for accounts.