Streamlined section - using equations to find the dimensions

[super sky]

Homework Statement

I need to find the values of thickness (t), mean diameter (D) and mean length (L) that will give the minimum area for the streamlined section, given that it's moment of inertia is 5.42 in^4.

Homework Equations

(the I's and A's after all K's here are meant to be subscripts, but they didn't appear properly...)
I = K$$_{I}$$tD$$^{3}$$
K$$_{I}$$ = 0.290R+0.054

A = K$$_{A}$$tD
K$$_{A}$$ = 1.875R+0.992

R = L/D

The Attempt at a Solution

Substitute K$$_{I}$$, K$$_{A}$$, R into both equations.

I thought I'd try using a MATLAB script (attached) to run through the different values of L,D,t but I don't know how to deal with 3D equations. Also, somewhere along the way I need to use the I = 5.42, but I don't know how to use that piece of information. So instead I solved the I equation for t using that I, and substituted that into the A equation. Then, making sure that t can't be greater than or equal to D/2 (since physically, that'd mean that the thickness is bigger than the radius which is obviously ridiculous). There's also a consideration that the formulae may or may not apply when the thickness to total length ratio is greater than 1/10, so I put that in there at the end but it didn't improve things. I ran the script and did a surface plot of the area, but I don't think I'm getting a sensible plot.

After that, I'm not quite sure how I'd find my values of t, D, and L... But as a first step, getting my solution in the right form would be the priority. It'd be great if anyone could help me out.

 

[nilesh_pat]

Thank you!<code>R = zeros(1,50);t = zeros(1,50);A = zeros(1,50);D = zeros(1,50);L = zeros(1,50);K_I = 0.290 * R + 0.054;K_A = 1.875 * R + 0.992;I = 5.42;for i = 1:50 L(i) = i; D(i) = i;endfor i = 1:50 R(i) = L(i)/D(i);endfor i = 1:50 t(i) = (I / (K_I(i)*D(i)^3)); if t(i) &gt;= D(i)/2 t(i) = D(i)/2; end A(i) = K_A(i) * t(i) * D(i); if R(i) &lt; 0.1 A(i) = 0; endend[L, D] = meshgrid(L, D);surface(L, D, A);</code>A:You can solve this problem using Calculus of Variations. In this case you have three independent variables (t, D and L) and you want the minimum of an expression which depends on them. The trick is to make a function of a single variable, say u, such that it is the expression you want to minimize, but this expression includes the three independent variables. So you define u as a function of t, D and L, sayu = f(t, D, L)You then take the derivatives of this with respect to t, D and L:du/dt, du/dD, du/dLYou write a set of equations using these derivatives which must be zero for the function u to have its minimum value. This gives you 3 equations in the 3 unknowns t, D and L.As an example, suppose that you wanted to find the minimum value of the expression xy. You would define u asu = xyThen, by taking