min max problems (1 Viewer)

darshil · May 4, 2009

Hey everyone,
We recently looked at min/max problems in geo application of calculus. Boy is it interesting, but at most of the questions I just cannot make it through, i just get stuck. What are some good ways to tackle these questions?

Here are some that get me stuck:

1) A traveller employs a man to drive him from Sydney to Wollongog for an hourly payment of P dollars. Running cost of the car, which are also paid by the traveller, are kv^3 per hour where v km/h is the speed and k is the constant. find the uniform speed that will minimze the total cost of the journey.

2) A company manufactures items at $2 per item and sells them at $x per item. If the number sold is 800/x^2 per month, find the value of x for which the company would maximise its monthly profits.

Thanks for all the help, I am forever indebted.

AnandDNA · May 4, 2009

[SOLVED] minimums - Math Help Forum

darshil · May 5, 2009

thanks for the reply man, it helped heaps

kurt.physics · May 5, 2009

darshil said:
2) A company manufactures items at $2 per item and sells them at $x per item. If the number sold is 800/x^2 per month, find the value of x for which the company would maximise its monthly profits.

$$The profit will be$\,\,$(x - 2)$

$Therefore, the total amount of profit per month will be:$

$(\frac{800}{x^2})\times(x-2)$

$= 800x^{-1} -1600x^{-2}$

$\frac{dP}{dx} = -800x^{-2} + 3200x^{-3}$

$$To find the maximum profit,$\,\, \frac{dP}{dx} = 0$

$$ie$\,\, -800x^{-2} + 3200x^{-3} = 0$

$$By multiplying by$\,\, x^3\,\,$we have:$

$-800x + 3200 = 0$

$800x = 3200$

$\therefore x = $4$

gurmies · May 5, 2009

kurt.physics said:
$$The profit will be$\,\,$(x - 2)$

$Therefore, the total amount of profit per month will be:$

$(\frac{800}{x^2})\times(x-2)$

$= 800x^{-1} -1600x^{-2}$

$\frac{dP}{dx} = -800x^{-2} + 3200x^{-3}$

$$To find the maximum profit,$\,\, \frac{dP}{dx} = 0$

$$ie$\,\, -800x^{-2} + 3200x^{-3} = 0$

$$By multiplying by$\,\, x^3\,\,$we have:$

$-800x + 3200 = 0$

$800x = 3200$

$\therefore x = $4$

Great work, but always remember to test that it is indeed a max by subbing into second derivative and appreciating that there is a -ve value.

GUSSSSSSSSSSSSS · May 5, 2009

回覆: Re: min max problems

yes i agree with gurmies, good solution, but MUST make sure to confirm it is a maximum, and also in many cases (particularly ones with restricted domains) you must test boundary conditions, ie: in the question above x coulda equalled, 0 or infinity, and by subbing those in you find that they produce results SMALLER than ur maximum and so can be disregarded xD

jet · May 5, 2009

Re: 回覆: Re: min max problems

gurmies said:
Great work, but always remember to test that it is indeed a max by subbing into second derivative and appreciating that there is a -ve value.

GUSSSSSSSSSSSSS said:
yes i agree with gurmies, good solution, but MUST make sure to confirm it is a maximum, and also in many cases (particularly ones with restricted domains) you must test boundary conditions, ie: in the question above x coulda equalled, 0 or infinity, and by subbing those in you find that they produce results SMALLER than ur maximum and so can be disregarded xD

Exactly. You would lose marks for not checking it in the HSC.

kurt.physics · May 5, 2009

LOL, i completely forgot ;P

PC · May 5, 2009

• Introduce the variables.
• Write down an equation connecting them.
• Obtain an expression for Q, the quantity you want to maximise or minimise.
• Express Q in terms of one variable using the equation from step 2.
• Differentiate and apply tests.

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