A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour. The total cost of the journey, $C$, is given by $$C = \frac{1400}{v} + \frac{2v}{7}.$$ (a) Find the value of $v$ for which $C$ is a minimum. (b) Find $\frac{d^2C}{dv^2}$ and hence verify that $C$ is a minimum for this value of $v$. (c) Calculate the minimum total cost of the journey.

A-Level Edexcel Maths Pure Proof: A diesel lorry is driven from Bi

A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

Question 10

A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour. The total cost of the journey, $C$, is given by $$C = \frac{1400}{v} ... show full transcript

Worked Solution & Example Answer:A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

Step 1

Find the value of $v$ for which $C$ is a minimum.

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Answer

To find the value of $v$ that minimizes $C$ , we first need to differentiate $C$ with respect to $v$ .

The function is given by:
$C = \frac{1400}{v} + \frac{2v}{7}.$

Differentiating with respect to $v$ :
$\frac{dC}{dv} = -\frac{1400}{v^2} + \frac{2}{7}.$

Setting the derivative equal to zero for minimization:
$-\frac{1400}{v^2} + \frac{2}{7} = 0.$

Rearranging gives:
$\frac{2}{7} = \frac{1400}{v^2}.$

Multiplying both sides by $7v^2$ :
$2v^2 = 9800.$

Dividing by 2:
$v^2 = 4900.$

Taking the square root:
$v = 70.$

Thus, the value of $v$ that minimizes $C$ is $70$ km/h.

Step 2

Find $\frac{d^2C}{dv^2}$ and hence verify that $C$ is a minimum for this value of $v$.

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Answer

Next, we need to find the second derivative $\frac{d^2C}{dv^2}$ .

Starting from the first derivative:
$\frac{dC}{dv} = -\frac{1400}{v^2} + \frac{2}{7}.$

Differentiating again:
$\frac{d^2C}{dv^2} = \frac{2800}{v^3}.$

Now, substituting $v = 70$ into the second derivative:
$\frac{d^2C}{dv^2} = \frac{2800}{70^3} = \frac{2800}{343000} \approx 0.0081 > 0.$

Since $\frac{d^2C}{dv^2} > 0$ , this confirms that $C$ is indeed a minimum at $v = 70$ .

Step 3

Calculate the minimum total cost of the journey.

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Answer

Finally, we find the minimum total cost $C$ by substituting $v = 70$ back into the cost function:

$C = \frac{1400}{70} + \frac{2(70)}{7}.$

Calculating each term:
$C = 20 + 20 = 40.$

Thus, the minimum total cost of the journey is $C = 40$ .

A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

Worked Solution & Example Answer:A diesel lorry is driven from Birmingham to Bury at a steady speed of $v$ kilometres per hour - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

Find the value of $v$ for which $C$ is a minimum.

Find $\frac{d^2C}{dv^2}$ and hence verify that $C$ is a minimum for this value of $v$.

Calculate the minimum total cost of the journey.

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