1. ## differential beams

Hey i have a question im getting no where with:

The strength of a rectangular beam of given length is proportional to bd3where b is the breadth and d the depth. If the cross section of the beam has a perimeter of 4m, find the breadth and depth of the strongest beam.

Thanks.

2. ## Re: differential beams

Strength s = kbd^3 for some constant k, and the perimeter condition enables you to express d in terms of b. So you have s as a function of b, and you are looking for the value(s) of b that maximises s...

Do you know what to do?

3. ## Re: differential beams

i understand the kbd^3 bit but the rest of what you mentioned i dont understand

4. ## Re: differential beams

So the context isn't exercises in finding maximum and minimum points (turning points) on the curve of a function? (By finding where the derivative = 0 ?) You haven't been doing exercises of that sort?

Do you mean you can't get d in terms of b? But the length of the perimeter is 4 = 2b + 2d...

5. ## Re: differential beams

i got that as the perimeter its just i was told by a friend that this has something to do with parametric equations, and yes i have been doing exercises of that sort. But in my revision notes I have no worked examples of a problem like this hence why im stuck and i have a mid term for this module next week.

6. ## Re: differential beams

Ok, it could be an exercise in Lagrange multiplier - Wikipedia, the free encyclopedia.

That way you don't eliminate b or d at the beginning, but find critical points of the two-variable function s = kbd^3, given the constraint, 2b + 2d = 4.

Same thing in the end, of course. You could try plugging the function and constraint straight into the formula (on the wiki page), but your best chance of understanding is to revise the basics by optimising a function of b as I suggested.

I.e. begin by making d the subject of 2b + 2d = 4. Get s in terms of k and b. Differentiate with respect to b, set the derivative equal to zero. That's where the gradient of the curve is zero and s is therefore a max or min.

7. ## Re: differential beams

you sure its about lagrange multiplier as thats not on our topic list, the topic list is

Differentiation of trig and log functions, product, quotient and function of function rules. Higher derivatives.

Hyperbolic and inverse hyperbolic functions and derivatives. Log equivalents of hyperbolic functions. Differentiation of inverse hyperbolic and inverse trig functions.

Applications of differentiation:Tangents speed and acceleration. Rates of Change. Local maxima and minima and applications

implicit and parametric differentiation. Logarithmic differentiation

partial differentiation

8. ## Re: differential beams

Originally Posted by tom@ballooncalculus
So the context isn't exercises in finding maximum and minimum points (turning points) on the curve of a function? (By finding where the derivative = 0 ?) You haven't been doing exercises of that sort?
Originally Posted by sheikh
Local maxima and minima and applications
Originally Posted by tom@ballooncalculus
I.e. begin by making d the subject of 2b + 2d = 4. Get s in terms of k and b. Differentiate with respect to b, set the derivative equal to zero. That's where the gradient of the curve is zero and s is therefore a max or min.
Maxima and minima - Wikipedia, the free encyclopedia

9. ## Re: differential beams

Thanks for the help, I got an answer of 1.69 to 2 d.p which I believe is correct.

Thanks again for the help

10. ## Re: differential beams

Originally Posted by sheikh
which I believe is correct.
How so?