Question

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
How much wire should be used for the square in order to minimize the total area?

Answer 1

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let t be the length of the wire used for the triangle. \\Let A_{S} be the area of the square. \\Let {A}_{T}} be the area of the triangle. \\One side of the square is \frac{s}{4} \\Therefore,we know that, A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, A=\frac{\sqrt{3}}{4} a^{2} where a is the length of one side,And one side of our triangle is \frac{t}{3} So,We know that, A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2} We have to find the value of "s" such that, \mathrm{s}+\mathrm{t}=19 hence, \mathrm{t}=19-\mathrm{s} And A_{S}+A_{T}=A_{S+T}$

$Therefore, \begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get, A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18} Now we solve A_{S+T}^{\prime}=0 \begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned} Cross multiply, \begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of s is [0,19] .So the endpoints are 0 and 19 \begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$Therefore, for the minimum area, 8.26 \mathrm{~m} should be used for the square$