Skip to content
3.1 · Q6

Q.The marginal cost of producing xx units of a product is given by MC=xx+1MC = x\sqrt{x+1}. The cost of producing 3 units is ₹7800. Find the cost function.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
88% · 52/59 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 →

Integrate MC=xx+1MC=x\sqrt{x+1} (put u=x+1u=x+1), then use C(3)=₹7800C(3)=₹7800 to fix the constant; the integral part at x=3x=3 is only 11215≈₹7.47\tfrac{112}{15}\approx₹7.47, so the constant carries almost all of ₹7800.

Total cost C(x)=∫MC dxC(x)=\displaystyle\int MC\,dx, constant fixed by the given C(x1)=C1C(x_1)=C_1. Substitution: u=x+1, x=u−1, dx=duu=x+1,\ x=u-1,\ dx=du.

  1. C(x)=∫xx+1 dx=∫(u−1)u du=∫(u3/2−u1/2) du.C(x)=\displaystyle\int x\sqrt{x+1}\,dx=\int (u-1)\sqrt{u}\,du=\int\big(u^{3/2}-u^{1/2}\big)\,du.
  2. =25u5/2−23u3/2+C=25(x+1)5/2−23(x+1)3/2+C.=\dfrac{2}{5}u^{5/2}-\dfrac{2}{3}u^{3/2}+C=\dfrac{2}{5}(x+1)^{5/2}-\dfrac{2}{3}(x+1)^{3/2}+C.
  3. Apply C(3)=7800C(3)=7800: at x=3x=3, u=4u=4, so (4)5/2=32, (4)3/2=8(4)^{5/2}=32,\ (4)^{3/2}=8.
  4. Integral part =25(32)−23(8)=645−163=192−8015=11215≈7.467.=\dfrac{2}{5}(32)-\dfrac{2}{3}(8)=\dfrac{64}{5}-\dfrac{16}{3}=\dfrac{192-80}{15}=\dfrac{112}{15}\approx 7.467.
  5. 11215+C=7800⇒C=7800−11215=117000−11215=11688815≈7792.53.\dfrac{112}{15}+C=7800\Rightarrow C=7800-\dfrac{112}{15}=\dfrac{117000-112}{15}=\dfrac{116888}{15}\approx 7792.53.
  6. Hence C(x)=25(x+1)5/2−23(x+1)3/2+11688815.C(x)=\dfrac{2}{5}(x+1)^{5/2}-\dfrac{2}{3}(x+1)^{3/2}+\dfrac{116888}{15}. …

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.