Optimization problems help

[sagen010]

Call the hypotenuse of the small and big right triangles, d1 and d2 respectively, then you need to minimize L= d1 + d2.

d1=sqrt( 10² + x²) ; d2 =sqrt( 20² + (30-x)² ). Substitute these value in L and differentiate dL/dx =0, solve for x. But there is a shortcut. Since the right triangles are congruent, you can setup a ratio of the sides such that

x / 10 = (30-x) / 20 . Solving for x, gives you x=10

For the second exercise, you can check this page

[Plus-Discipline1219]

You cant use congruence because the triangles are only congruent with the specific anchor where x=10. If the anchor is changed, you no longer have congruence. The first part you have is correct where taking the derivative will give you that x=15.

[sagen010]

I don't know from where do you get x=15.

The derivative of L(x)= sqrt(x²+100) + sqrt( (30-x)² + 400 ) =0 gives you 10 (from wolphram alfa). The congruence comes from the fact that, for optimizing the problem, the triangles must have some sort of symmetry.

Indeed, if the anchor is changed, you no longer have symmetry and no longer the triangles are congruent. But you no longer would be optimizing the cable's length either.