Maths here: >rectangle >periphery of 48 meters >blue

Call the sides of the full rectangle x and y then the problem becomes:

max x*y - 2*(y/2)^2
s.t. 2x+2y = 48

You could either use the method of Lagrange multipliers or you could use the constraint to remove a variable and look where the derivative is 0.

>>207407
It should be doable with a simple quadratic equation. I do already know the solution (it's 16 and 8, x should be 2y), I'm just not sure how to get there.

>>207410
Well, If we calculate the zero-values of the red surface area (which is A = -6y2 + 48y), we get 0 and 8. This would make the other side 16.

That about right?

>>207405
Easier solution: Move both of the blue squares to one side. The areas and perimeters remain the same.

Let y <= x. Then you need to max y(x - y/2) such that 2x + 2y = 48, i.e., x = 24 - y.

y(x - y/2) = y(24-3y/2) = (3/2)y(16 - y)

The zeros are at y = 0 , y = 16, the maximum is attained at y = (16-0)/2 = 8, x = 24 - 8 = 16.
The maximum area is 8(16-4) = 96.

>>207441
Thanks!

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Hi homework request. I've been given what looks like a reimann sum and told to write the definite integral. I don't understand how to answer this question. I noticed that delta x is 3/n (from (5-2)/n) so I am left with the function times delta x. This is the part of the problem I need help with.

I think I know how to do the second part ( I am attempting it now) (I think I need to find the function without absolute values and take the different definite integrals, graph it and find the areas.)

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Homework question here, I have to find 4 positive integers a, b, c, and n so that a^n + b^n = c^n for some value of n greater than 2. Need it by tomorrow. Help.

>>207690
There's a really elegant solution to this problem, but it's a little bit too long to fit in the comment box.