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 + 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.
3272 = 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=27
Doing it a step at a time:
Square both sides, so x3=272, then cube root both sides, so 3272=x
or in one step, raise each side to the power of the reciprical of the exponent (i.e. 2/3)
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 + 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.
3272 = 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 x23=27
You want this in the form x=…
Using the rules of indices, when you raise a something with a power to another power, you multiply the indices. eg (a2)21=a2×21=a.
Therefore, in this question, you can raise both sides of the equation to the pwoer of 32