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3.5 · Q8

Q.The cost of fuel in running an engine is proportional to the square of the speed in kms per hour, and is ₹48 per hour when the speed is 16 km. Other costs amount to ₹300 per hour. Find the most economical speed.

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Fuel cost per hour is 316v2\tfrac{3}{16}v^2; minimising the total cost per km 316v+300v\tfrac{3}{16}v+\tfrac{300}{v} gives the most economical speed v=40v=40 km/hr.

Cost per km C(v)=cost per hourspeed=kv2+(fixed cost/hr)vC(v)=\dfrac{\text{cost per hour}}{\text{speed}}=\dfrac{kv^2+\text{(fixed cost/hr)}}{v}. Minimise via C′(v)=0C'(v)=0, C′′(v)>0C''(v)>0.

  1. Fuel cost per hour ∝v2\propto v^2, i.e. =kv2=kv^2. Given ₹48 per hour at v=16v=16: k(16)2=48⇒256k=48⇒k=48256=316k(16)^2=48\Rightarrow 256k=48\Rightarrow k=\dfrac{48}{256}=\dfrac{3}{16}.
  2. Other costs =₹300=₹300 per hour, so total cost per hour =316v2+300=\dfrac{3}{16}v^2+300.
  3. Cost per kilometre (divide by speed vv): C(v)=316v2+300v=316v+300vC(v)=\dfrac{\tfrac{3}{16}v^2+300}{v}=\dfrac{3}{16}v+\dfrac{300}{v}.
  4. Differentiate: C′(v)=316−300v2C'(v)=\dfrac{3}{16}-\dfrac{300}{v^2}. …

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