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Precalculus

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There is a piece of wire 200 cm long. It needs to be cut into 2 pieces, with one piece forming an equilateral triangle and the other piece forming a circle. How long must each piece be to maximize the area of each shape?

  • Precalculus -

    area= PI*(L/2)^2 + area of triangle.

    Well, it turns out there is a neat formula

    area= sqrt(s(s-a)(s-b)(s-c)) where s is the semi perimeter, and a,b,c are sides.

    s= 1/2 (3T) where T is triangle side
    so areatriangle= sqrt(3T/2)(3T/2-T)^3
    = sqrt(3T/2 * T^3/8)=sqrt(3T^4/16)
    = T^2/4 * sqrt3

    so finally, area total=

    PI (L/2)^2+ T^2/4 sqrt3
    where 3T+PiL= 200cm

    or Area=1/4PI* (200-3T)^2+T^2/4 sqrt4
    check that.
    Now with calculus, you take the derivative of area, set to zero, and solve for T, then L.
    I assume you have not had this, then plot the parabola, find its maximum, and the corresponding T.

    Knowing T, cut three pieces T length, and the remainder Pi*L is the circle perimeter.

    check my work.

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