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(edited 6 years ago)
Original post by MathsAssist
I am having some trouble understanding the operations used to solve this question:

A steam train travels between Haverthwaite and Eskdale at a speed of x miles per hour and burns y units of coal, where y is given by: 2x\sqrt x + 27/x for x > 2.

a) Find the speed of that gives the minimum coal consumption.

To solve this, I did dy/dx, then then dy/dx = 0:

Simplifying the formula: 2x^1/2 + 27x^-1

dy/dx = 2x^-1/2 - 27x^-2
0 = 2x^-1/2 - 27x^-2
27/x^2 = 1/√x
27 = x^2/x^1/2
27 = x^3/2 This is the last bit I managed to do on my own.

2372\sqrt[3]27^2 = x (This is the part I don't understand - how did they get to this from 27 = x^3/2)

so x = 9


x3/2=27x^{3/2}=27

Doing it a step at a time:

Square both sides, so x3=272x^3=27^2, then cube root both sides, so 2723=x\sqrt[3]{27^2}=x

or in one step, raise each side to the power of the reciprical of the exponent (i.e. 2/3)

(x3/2)2/3=x3/2×2/3=x=272/3=2723\displaystyle \left(x^{3/2}\right)^{2/3}=x^{3/2\times 2/3}=x=27^{2/3}=\sqrt[3]{27^2}
(edited 6 years ago)
Original post by MathsAssist
I am having some trouble understanding the operations used to solve this question:

A steam train travels between Haverthwaite and Eskdale at a speed of x miles per hour and burns y units of coal, where y is given by: 2x\sqrt x + 27/x for x > 2.

a) Find the speed of that gives the minimum coal consumption.

To solve this, I did dy/dx, then then dy/dx = 0:

Simplifying the formula: 2x^1/2 + 27x^-1

dy/dx = 2x^-1/2 - 27x^-2
0 = 2x^-1/2 - 27x^-2
27/x^2 = 1/√x
27 = x^2/x^1/2
27 = x^3/2 This is the last bit I managed to do on my own.

2372\sqrt[3]27^2 = x (This is the part I don't understand - how did they get to this from 27 = x^3/2)

so x = 9


You have x32=27x^{\frac{3}{2}}=27

You want this in the form x=x= \dots

Using the rules of indices, when you raise a something with a power to another power, you multiply the indices. eg (a2)12=a2×12=a(a^2)^{\frac{1}{2}} = a^{2\times \frac{1}{2}} = a.

Therefore, in this question, you can raise both sides of the equation to the pwoer of 23\frac{2}{3}

(x32)23=2723(x^{\frac{3}{2}})^{\frac{2}{3}}=27^{\frac{2}{3}}

x=2723\Rightarrow x = 27^{\frac{2}{3}}, since 32×23=1\frac{3}{2} \times \frac{2}{3} = 1 as the power of xx.

You should know that 2723(273)227^{\frac{2}{3}} \equiv (\sqrt[3]{27})^2

Hope that helped :smile:
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(edited 6 years ago)
Original post by MathsAssist

do you know any good places I can find complex-ish questions involving rules of indices? I think I need some more practice with them before I'm solid.


Not familiar with what's out there on the 'net, so can't help on that one.

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