maths help for as-level?

I will assume that the equation is S=(160/(x^2))+(5*x) since it is written in a way that could also indicate 160/((x^2) + 5*x)...

The correct way of solving both a and b is to begin with taking the derrivative of S with respect to x:

Sometimes the derivative of S is written S', I will use a different notation where the derivative of S with respect to x is written dS/dx or d/dx (S).

dS/dx = d/dx ((160/(x^2))+(5*x)) = d/dx ((160*(x^(-2))+(5*x)).

The last part is simple, d/dx (5*x) = 5.

So dS/dx = 5 + d/dx (160*(x^(-2)))

= 5 + 160 * d/dx (x^(-2))

Now use that d/dx (x^n) = n*x^(n-1):

then d/dx (x^(-2)) = -2*x^(-3).

So dS/dx = 5 + 160 * (-2)*x^(-3)

= 5 - (320 / x^3)

To find the minimum, we need to establish where the function S is flat, meaning the derivative is zero.

dS/dx = 0 <=> 5 - (320 / x^3) = 0

<=> - (320 / x^3) = -5

<=> 320 / x^3 = 5

<=> 320 / 5 = x^3

<=> 64 = x^3

<=> 3rd root (64) = x

<=> 4 = x.

This means that 4 is the only flat point on the function, and therefore the only possible location of a minimum.

To check that it actually is a minimum (could be a maximum or a "turning tangent"), calculate the value of 4 and of a few points on either side of 4:

S(x=4) = (160/(4^2))+(5*4) = 160/16 + 20 = 10+20 = 30.

S(x=3) = (160/(3^2))+(5*3) = 160/9 + 15 = 17.78 + 15 = 32.78.

S(x=5) = (160/(5^2))+(5*5) = 160/25 + 25 = 6.4 + 25 = 31.4.

It sure looks like 4 is a minimum.

To be absolutely sure, you could check some values even closer to 4, like 4.01 or 3.99.

f'(x) = -32x^-3 + 5

Stationary point x = 1.8566

(this is found by equating f'(x) to 0)

f''(x) = 96x^-4

Substitute the x value here and you'll get 8.0797. Since this is greater than 0, the value is a minimum.