3)
You plan to make a simple, open topped box from a piece of sheet metal by cutting a square โ of equal size โ from each corner and folding up the sides
If ๐ = 200๐๐ and ๐ค = 150๐๐ calculate:
a) The value of x which will give the maximum volume
b) The maximum volume of the box
c) Comment of the value obtained in part b.
4) The gain of an amplifier is found to be
๐ฎ = ๐๐ ๐ฅ๐จ๐ (๐๐๐ฝ๐๐๐):
The tasks are to find equations for:
a) ๐๐บ/๐๐๐๐ข๐ก
b) ๐^2๐บ/๐๐๐๐ข๐ก 2
3.
The volume of the box can be written in the form:ย "V(x)=(L - 2\\cdot x)\\cdot(W- 2\\cdot x)\\cdot x"
Lengths and width of the box decreased that is of sheet metal byย "x"ย from each corner, and height of the box is equalย "x". We bringย "V(x)"ย to a simple form:ย "V(x)=4\\cdot x^3 -2\\cdot (L+W)\\cdot x^2+ L\\cdot W\\cdot x"
To find maximum volume one compute the derivative of volume with respect toย "x"
"V^{'}_x=12\\cdot x^2 - 4\\cdot (L+W)\\cdot x+ L\\cdot W"ย and define the root of the equationย "V^{'}_x=0"ย :
"x_{1,2}=(2\\cdot(L+W)\\pm\\sqrt{4(L+W)^2-12\\cdot L\\cdot W} )\/12=\\frac{1}{6}(L+W\\pm\\sqrt{L^2+W^2-L\\cdot W})"
"x_1=88.38 ;\\space x_2=28.29"
The first value cannot be implemented. It is clear that the box will succeed only ifย "x<W\/2"
The second value corresponds to the maximum volume shown in the figure.
Answer:ย "x=28.29 mm; max V=379037.81 mm^2"
4.
"(a) d G \/ d V o u t=20 * 1 * 10 \/(10 * Vout * \\ln 10)=20 \/( Vout * \\ln 10)\n\n\\\\(b)\nd^{2} G \/ d V o u t^{2}=(-20 * d(\\text { Vout } * \\ln 10) \/ d V \\text { out }) \/(\\text { Vout } * \\ln 10)^{2}\n\\\\=-20 * \n\\ln 10 \/\\left((\\ln 10)^{2} * \\text { Vout }^{2}\\right)\n=-20 \/\\left(\\ln 10 * V o u t^{2}\\right)"
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