Maths question ~

right ok, I'm trying to use your way to figure out the prism question.....so.....

DV = (sqrt 3)/4 b^2 dx + (sqrt of 3)/2 bx db = 0
DA = (sqrt of 3) b db + 3b dx + 3x db = 0

Is that right? o_O

So far so good

now set the dependant variables to zero (0) for their maxima and minima, and separate db and dx into factor groups (hint the volume differential is easy, the surface area has db in two terms.

then you'll see two different derivative functions of the form g'dx+f'db=0 which can be rearranged to the form dx/db=-f'/g' (or db/dx=-g'/f'). set each equal to each other and solve for one in terms of the other. This is your optimum, that maximizes the volume of the cylinder of this shape for the surface area.

Neat huh!--

Rap

one for volume, the other for surface area

ahh ok, i hate working wit square roots so I'm getting confused:

So in DV, the common factor I take out is ( (sqrt 3)/2 )*b to give:

( (sqrt 3)/2 )*b [ (1/2) * b dx + x db ]

which gives:

1/2 b dx = -x db
dx/db = -2x/b ( ?? Is that right?)

and I'm totally confused with the area differential - am I suppose to take out a common factor from the dx terms first?

Also I was looking through my textbook and I came across a similar differential question which requires minimisation.

An oil production platform is 9/(sqrt 3) km directly offshore from nearest pt A, is to be connected by a pipeline to an onshore refinery R, 100km down from the coast along the straight coatline from A. It costs £2 million per km to lay it underwater, £1 million per km to lay pipeline onshore. By introducing one unknown, write an expression for the cost of pipeline. Find the minimum value of this cost.

Hint - Let x be the unknown and y to be the cost. Set dy/dx = 0 to find the minimum.

I know how to come about solving this for dy/dx but I don't know what expression to start off with. Any suggestions?

good on dV

now look at Surface area

S=sqrt(3)/2b^2+3bx

when you differentiate this the term (3bx) has both independent variables so you gotta use g'db+f'dx on this one and you'll end up with a db in two places. Don't worry you can separate them by using the rules if distribution you learned in algebra.

Then you'll be able to set dx/db just like you did with the volume expression. Equate them and solve for the optimum b in terms of x, or x in terms of b.

Now the pipeline problem.

Draw a right triangle with one leg equal to 100km and the other 9/sqrt(3) km. Pick a point on the longer leg, and create a second right triangle (inside the first) with the hypotenuse between this point and the platform.

If you want labels, label the hypotenuse of the second triangle y and the height x (you already know the base its 9/sqrt(3)km). Think about Pythagoras and how he would have expressed y in terms of x.

Now y (in terms of x) is the sea leg of the pipeline, and the land leg is then (100-x), and the total cost is then

C=3y+2(100-x)

differentiate cost with respect to x and set to zero to minimize.

Remember to substitute y in terms of x before you differentiate.

Rap