Find dimensions of the rectangular box of maximum volume that can be inscribed in a sphere of radius a?

1 answer:

4 0

This is an optimization calculus problem where you would need to know a little bit more about the box, atleast i would think. You would just need to use the volume equation of a sphere as the restrictive equation in the optimization problem. Perhaps there is a way to solve with the given information, but i do not know how to.

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Use each integer only once. You must have each row, column, and diagonal adding up to the magic sum.

Grace [21]

For question number 3
-7, 7, 6, -4,
4, -2, -1, 1
0, 2, 3, -3,
5, -5, -6, 8

For question number 4:
-4, 1, -10, 3,
-9, 2, -3, 0,
5, -8, -1, -6,
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8 0

2 years ago

Read 2 more answers

Solve for x, 0 ≤ x ≤2π (2cosx-1)(2sinx+√3 ) = 0

vladimir1956 [14]

Answer:

x = π/3, x = 5π/3, x = 4π/3

Step-by-step explanation:

Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3 = 0.

$2\cos \left(x\right)-1=0,\\2\cos \left(x\right)=1,\\\cos \left(x\right)=\frac{1}{2}$

Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.

Similarly:

$\:2\sin \left(x\right)+\sqrt{3}=0,\\2\sin \left(x\right)=-\sqrt{3},\\\sin \left(x\right)=-\frac{\sqrt{3}}{2}$