Answer:
Dimensions: 0.4 by 3.2 by 1.2
Step-by-step explanation:
I'm assuming here that we are cutting out squares out of each of the metal's corners:
Let x = the length of each cut-out square,
Each base (of the desired net square folded) is 4-2x, and 2-2x respectively,
Volume = x(4-2x)(2-2x)
= 4x^3 - 12x + 8x
Now we take the derivative:
[tex]\frac{d}{dx}\left[4x^3-12x^2+8x\right]\\\\= \frac{d}{dx}\left(4x^3\right)-\frac{d}{dx}\left(12x^2\right)+\frac{d}{dx}\left(8x\right)\\\\= 12x^2-24x+8[/tex]
We equate to 0 (0 for max volume), and solve using the quadratic formula:
[tex]12x^2-24x+8=0,\\\\x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:12\cdot \:8}}{2\cdot \:12}\\\\= \frac{-\left(-24\right)\pm \:8\sqrt{3}}{2\cdot \:12}\\\\\mathrm{Separate\:the\:solutions}:\\\\x_1=\frac{-\left(-24\right)+8\sqrt{3}}{2\cdot \:12},\:x_2=\frac{-\left(-24\right)-8\sqrt{3}}{2\cdot \:12}\\\\x =\frac{3+\sqrt{3}}{3},\:x=\frac{3-\sqrt{3}}{3}\\\\x=1.57735\dots ,\:x=0.42264\dots[/tex]
So we approximate the side lengths to be 1.6 and 0.4 respectively. But when we plug in 1.6 for x, we get the volume as negative. Therefore x has to be 0.4.
Side lengths: 0.4, 4-2(0.4) => 3.2, 2-2(0.4) => 1.2
