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I do not know how to deal with these problems. Http: / / www.cet.ac.il/math/function/english/square/multi/multi7.htm Can anyone help? Is there any way derived free to do so? I did not know that knowledge of algebra II requires derivatives.
In each issue as that has two equations. An equation is on the perimeter of the figure and the other will be for the region. You will solve for the perimeter of a variable and plug in the field of the equation. You then vary the field equation, set the derivative equal to zero and solve. This value can be computed the other dimension. For the first example is the "duration" (the horizontal dimension to the place where the cycle begins) and L is "wide" (the vertical dimension) is w. Then the equation for the perimeter, you have: 40 = 2L + W + πw / 2 = 2L W + (1 + π / 2) πw / 2 occurs because circumeference a complete circle is π times the diameter and the diameter of the semicircle on the right is W. But since this is a semi-circle, the distance is about π2 / 2. The area is the area of the rectangle (length by width or LW) and half of a circle (1 / 2 π (w / 2) ²) = π / 8 ² O). It would seem easier to solve the equation of the perimeter of the l: 40 – w (1 + π / 2) = 2L 2L = 40 – F (1 + π / 2) L = 20 – W / 2 (1 + π / 2) = 20 – F (1 / 2 + π / 4) LW = + π / 8 ² replace w: A (w) = (20 – W (1 / 2 + π / 4)) w + π / 8 W ² A (w) = 20W – (1 / 2 + π / 4 – π / 
w ² A (w) = 20W – (1 / 2 + π / w ² A (w) = 20W – (4 + π) / 8 w ² differentiate: A '(w) = 20 – (4 + π) / 4 W = 0 (4 + π) / 4 W = 20 W = 80 / (4 + π) ≈ 11.20 l = 20 – 80 / (4 + π) * (2 + π) / 4 l = 20 20 (2 + π) / (4 + π) ≈ 5.6 that took a while … not 100% sure of my work … but the way they must attack them all.
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