Answer:
See below for all the cube roots
Step-by-step explanation:
<u>DeMoivre's Theorem</u>
Let
be a complex number in polar form, where
is an integer and
. If
, then
.
<u>Nth Root of a Complex Number</u>
If
is any positive integer, the nth roots of
are given by
where the nth roots are found with the formulas:
for degrees (the one applicable to this problem)
for radians
for 
<u>Calculation</u>
<u />![z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)](https://tex.z-dn.net/?f=z%3D27%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%5C%5C%5C%5C%5Csqrt%5B3%5D%7Bz%7D%20%3D%5Csqrt%5B3%5D%7B27%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%7D%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D%2827%28cos330%5E%5Ccirc%2Bisin330%5E%5Ccirc%29%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D27%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%28cos%28%5Cfrac%7B1%7D%7B3%7D%5Ccdot330%5E%5Ccirc%29%2Bisin%28%5Cfrac%7B1%7D%7B3%7D%5Ccdot330%5E%5Ccirc%29%29%5C%5C%5C%5Cz%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%3D3%28cos110%5E%5Ccirc%2Bisin110%5E%5Ccirc%29)
<u>First cube root where k=2</u>
<u />![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%282%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B720%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B1050%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28350%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28350%5E%5Ccirc%29%2Bisin%28350%5E%5Ccirc%29%5Cbiggr%5D)
<u>Second cube root where k=1</u>
![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%281%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B690%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28230%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28230%5E%5Ccirc%29%2Bisin%28230%5E%5Ccirc%29%5Cbiggr%5D)
<u>Third cube root where k=0</u>
<u />![\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27%7D%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%2B360%5E%5Ccirc%280%29%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28%5Cfrac%7B330%5E%5Ccirc%7D%7B3%7D%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcis%28110%5E%5Ccirc%29%5Cbiggr%5D%5C%5C3%5Cbiggr%5Bcos%28110%5E%5Ccirc%29%2Bisin%28110%5E%5Ccirc%29%5Cbiggr%5D)
The volume as a function of the location of that vertex is
... v(x, y, z) = x·y·z = x·y·(100-x²-y²)
This function is symmetrical in x and y, so will be a maximum when x=y. That is, you wish to maximize the function
... v(x) = x²(100 -2x²) = 2x²(50-x²)
This is a quadratic in x² that has zeros at x²=0 and x²=50. It will have a maximum halfway between those zeros, at x²=25. That maximum volume is
... v(5) = 2·25·(50-25) = 1250
The maximum volume of the box is 1250 cubic units.
Answer:
Below.
Step-by-step explanation:
30 kg 500g + 28kg 700g
= 58kg + 1200g
= 59kg 200g.
Difference
= 30 kg 500g - 28kg 700g
= 2kg + 500 - 700
= 2kg - 200g
= 1kg 800g.