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M2007U-E2016-A2017-S2017-M2021_SMT-20200425x01
Let's draw Right Triangle with a Height of Th, Width with Tw, and install a rectangle inside it.
In this article, we are going to prove the largest rectangle we can make in this triangle.
Let's say the Height of the Triangle is U, and the width of the triangle is V
Let's try to find the relation between H and W
∵ W // X, H // Y
∴ α = β
∴ tanα = tanβ
Y / W = H / X
H = XY / W
H = (V - W)(U - H) / W
HW = (V - W)(U - H)
HW = VU - VH - UW + HW
0 = VU - VH - UW
VH = VU - UW
H = U - UW/V
∴ the area of the rectangle would be :
A = WH
A = W(U - UW/V)
A = (UW) - (UW²)/V
A = (U)W - (U/V)W²
so we can treat W as an input and the area A would be the output, both U and V are fixed constants
To make sure that the area of the rectangle has a maximum area, we just need to differentiate the equation twice and see if the result is negative or not.
dA/dW = U - 2(U/V)W
d²A/dW² = 0 - 2(U/V)
d²A/dW² = - 2(U/V)
since U and V are positive, then U/V > 0, then - 2(U/V) < 0, so the Area of the Rectangle A has a maximum value. When A is the maximum value, dA/dW = 0
∴ when dA/dW = U - 2(U/V)W = 0
1 - 2(1/V)W = 0
1 = 2(1/V)W
1/2 = W/V
W = (1/2)V
surprisingly, when W is half of the width of the triangle, the area of the rectangle is at its maximum, we and also substitute when A = Amax & W = (1/2)V to find that
H = U - UW/V
H = U - U((1/2)V)/V
H = U - (UV)/(2V) H = U - (1/2)U
H = (1/2)U
Did you know ?
This proof is inspired by the main task of filling up a Triangular region in Minecraft by using the command /fill in order to achieve the maximum efficiency
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