How to find the maximum in a function?

Question

i would like to know how to find the max. in a function

like the maximum of the function:

A(w)=w(40-2w/2.57)

how do u go about finding the max that "A(w)" can be.

thank you

Answer 1

The maxima will be a critical point where the surface is otherwise locally concave down.

Take the derivative and second derivative. Solve equations for zeros. Evaluate the function at the zeros (critical points) and look for the max. If you are given boundaries, check the boundaries AND critical points.

In terms of this function, its a lot easier if you multiply that w through, you see that it is a downward facing parabola, the y-intercept is at zero, so the maximum is on the right hand side plane.

The function is:
A(w)=40*w-(200/257)*w^2

The derivative is:
A'(w)=40-(400/257)*w

It evaluates to zero at 25.7 so thats w-location of the max

The value of the function at w=25.7 is 511.4786...

Answer 2

Make a rough sketch of the graph. This one is a downwards facing parabola.

Take the first derivative of A(w) with respect to w. Maxima and Minima will only occur at critical points, points where the first derivative = 0 or where the first derivative is undefined. Because in is a downwards facing parabola, we expect there to be only one critical point, which should correspond to the maximum value.
Here A(w) = 40w -(2/2.57)w^2
So dA/dw = 40 - (4/2.57)w, 40 - 1.56w (Power rule)
dA/dw is defined everywhere.
Set dA/dw = 0.
0 = 40 - 1.56w 1.56w = 40
The derivative has only one root value of an answer, and we guess that is the max.
w(max) = 40/1.56 = 25.7
Now plug w(max) into the funtion A(w) to find A(max)
A(max) = 40*25.7 - 2/2.57*(25.7^2) = 514.

Because there is only one critical point, there will be no more than one max. To make sure that (25.7, 514) is a max we simply need to test a point before w = 25.7 and a point after 25.7. If 97.66 is the maximum value of A(w), both test points should have a lower value of A(w). For w = 0, A(w) = 40*0 - (2/2.57)*0 = 0. 0< 514 So far so good. For w = 100, A(w) 40*100 - 20000/2.57, which is way less than 514. Thus 514 is the maximum value of A(max)

EDIT: I noticed that I accidentally misplaced the decimal point in w due to a transcription error. I then propagated this error throughout the problem. The corrected version appears here.

Answer 3

Generally, differentiate and find the turning points (where the first derivative equals zero).

The turning points may be maxima, minima or points of inflexion. You can check which they are by sketching the graph and using observation or ...

... you can differentiate again and see what value thus second derivative has at the point of interest. If the second derivative is negative then the point is a maximum, if the second derivative is positive then the point us a minimum, if the the second derivative is zero then the point is a point of inflexion.

In the example you have here
dA/dW=40 - 1.56w
At turning points dA/dW = 0
so 40 -1.56w=0
w=40/1.56
w=25.641 (approx) at the turning point.

when w=25.641
A(w) = 25.641(40-(2*25.641)/2.57)
A(w) = 514

Differentiating a second time

d²A/dW²= -1.56

So the second derivative is negative everywhere (including at the turning point) so the turning point is a maximum.

Summarising the maximum value of A(w) is 514,
when w = 25.641

Answer 4

transform A(w) so that we will have the form : constant - (....)^2 <= constant

A(w)=w(40-2w)/2.57 = 2 (20w - w^2) / 2.57
= 2 [ 100 - (w - 10)^2 ] / 2.57
= [200 - 2(w - 10)^2 ] / 2.57 <= 20000 / 257

so the max value of A(w) = 20000 / 257 when w = 2

Answer 5

The easiest way is to graph the function A(w).
The more mathmatical way is (normaly) to differenciate the function {find dA/dw}. Maxima and minima usualy occur at dA/dw = 0.