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Maximize the area of the rectangle that can be inscribed in a triangle with base 4 and height 3? Assume its isoceles..
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lagrange multipliers are useful for this, but i cant recall them in detail at the moment
Well there is a solution using single variable calculus.
one idea i gots is when the area of the trianlge is halved, we have an equilibrium point
4*3 - 12/2 = 6 when the area of the rectangle is 3; we are at the most that we can take in and the rest just makes our rectangle smaller in area
2 by 1.5 is my guess
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mines wrong .. i did a right triangle by mistake
For your problem the equations are \( A = 2xy \) and \( 4x + 3y = 12 \)
Now substitute x, or y and differentiate w.r.t to y or x.
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