Unit 3 optimisation- help plz

[Maz]

hey, i was just wandering if anyone could help me with this questions please

The cost, $C, for the production of x units of a certain product is given by
C=0.025(x)^2+2x+ 1000   ,  x is greater than 0
Use calculus to find the value of x for which the average cost per unit is a minimum, using the second derivative tests to confirm a minimum, and find this minimum average cost.

thankyou in advance

[zsteve]

Your cost for x units is C. Hence the average cost for each of those units is C/x.

We want to find x for which the avg. price is minimum, so find C'

Solve for dC/dx = 0 and that's your minimum. Solving gives
But x>0 so x = 200
Show that this is a minimum: find C''

At x = 200, C'' > 0.
This tells us that C' is increasing (as C'' > 0). But C'=0 at x=200 and increasing, so it goes from negative to positive.
Hence C' goes from - to + at x=200, so it's a minimum (confirmed)
Minimal average cost is C(200) = ...

[Maz]

thankyou so much

[tedescostudent]

Is double derivative something expected for a methods student? If so, may someone explain how it works?

[Swagadaktal]

Is double derivative something expected for a methods student? If so, may someone explain how it works?

no it isnt expected, however, in this years exam they did have a question where you could use double derivatives to answer it (im not sure if there is any other way to do it but there must be)
I'm going to leave a spesh student to answer the second part of your question, but yeah double derivatives are not covered

[Maz]

Is double derivative something expected for a methods student? If so, may someone explain how it works?

its expected in western australia- I'm doing 2nd derivative and doing min/max stuff with that-and I'm not even doing spec

[Gentoo]

The double derivative is to the derivative what the derivative is to the original function: it represents the RATE OF CHANGE of the derivative (like the derivative represents the RATE OF CHANGE of the original function). It is denoted by f''(x) and you calculate it by differentiating f'(x) (the derivative of f(x)).

The double derivative can be used (although of course it is never necessary in Methods) to verify the nature of stationary points, once you have used the derivative to find them.

For instance, if you find that for f(x), the derivative f'(x)=0 when x=3, x=5 and x=9, you can see that stationary points occur here. But in order to find out what type of stationary point they are, we need to either draw up a gradient table (this is what they teach you in Methods) OR calculate the double derivative at these points: f''(3), f''(5) and f''(9).

For instance, if f''(3) were to be positive, you could tell that the stationary point at x=3 is a local minimum, because while the gradient/derivative is zero at that point, the RATE OF CHANGE of the gradient (represented by the double derivative f''(3)) is positive. Intuitively, have a look at a local minimum turning point on any graph; see how the gradient starts off negative, then hits zero at the turning point, then becomes positive? That's because the rate of change of the gradient (the double derivative) is positive which causes the gradient to increase, and that's why we can use this test to determine that the stationary point is a local minimum. On the flipside, If f''(5) were negative, we could tell that there is a local maximum at x=5. If f''(9) were zero, there would be a stationary point of inflexion at x=9 (think of the point (0,0) on a x^3 graph).

Doing the double derivative test and drawing up a gradient table (to see what the gradient was before the stationary point and after) are two methods of determining what type of stationary point it was. You don't have to use the former in Methods.

[tedescostudent]

Hey Gentoo, so stationary point is at f'(x)=a for example.
Minimum: f''(a)>0
Maximum: f''(a)<0
Point of inflection: f''(a)=0
Is this correct? Would there be any other uses for double derivative for methods? Thanks for clarifying that with me Gentoo, appreciate it!

[zsteve]

Is double derivative something expected for a methods student? If so, may someone explain how it works?

its expected in western australia- I'm doing 2nd derivative and doing min/max stuff with that-and I'm not even doing spec

Methods students are assumed to be aware that it exists, but don't use it for anything. The second derivative method is accepted for methods exam but you have to interpret it as I did in my solution, i.e. 'dy/dx' is increasing so it goes from -ve to +ve or whatever.
In specialist you can make a statement about concavity instead.

[Gentoo]

Hey Gentoo, so stationary point is at f'(x)=a for example.
Minimum: f''(a)>0
Maximum: f''(a)<0
Point of inflection: f''(a)=0
Is this correct? Would there be any other uses for double derivative for methods? Thanks for clarifying that with me Gentoo, appreciate it!

Yeah, that's right.

The only other use for the double derivative in methods that I can think of is a question like: "when is the gradient of f(x) a minimum/maximum?" In this case, you would be solving f''(x)=0, just like how in order to find when f(x) is a maximum you would solve f'(x)=0.

[zsteve]

Yeah, that's right.

The only other use for the double derivative in methods that I can think of is a question like: "when is the gradient of f(x) a minimum/maximum?" In this case, you would be solving f''(x)=0, just like how in order to find when f(x) is a maximum you would solve f'(x)=0.

Yup, that's definitely on the course, turned up on 2015 VCAA MM CAS Exam 2

[Gentoo]

Ah yes, so it was. And it should remain on the new study design.