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Use lagrange multipliers to find the maximum volume of a rectangular box that is inscribed in a sphere of radius r

vesna_86 [32]

For the derivative tests method, assume that the sphere is centered at the origin, and consider the circular projection of the sphere onto the xy-plane. An inscribed rectangular box is uniquely determined

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by the xy-coordinate of its corner in the first octant, so we can compute the z coordinate of this corner by

x2+y2+z2=r2 =⇒z= r2−(x2+y2). Then the volume of a box with this coordinate for the corner is given by

V = (2x)(2y)(2z) = 8xy r2 − (x2 + y2),

and we need only maximize this on the domain x2 + y2 ≤ r2. Notice that the volume is zero on the boundary of this domain, so we need only consider critical points contained inside the domain in order to carry this optimization out.

For the method of Lagrange multipliers, we optimize V(x,y,z) = 8xyz subject to the constraint x2 + y2 + z2 = r2<span>. </span>

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3 years ago

an ordered pair where the x and y coordinate are the same lies in the 1st or 3rd quadrant. true, sometimes true or not true?. ex

Zarrin [17]

The quadrants are as follows:

The 1st quadrant has the points which have both x and y positive and the 3rd quadrant has the points which have both x and y negative. If the ordered pair and the same x and y value, if ons is positive, the other also is, and the same for negative.

So, at first we see that there are point where the x and y are the same and that are in the 1st or 3rd quadrant.

However, there is one special case:

When x and y are 0, that is, the ordered pair is (0, 0).

Since this point is the origin, it doesn't lie on any of the quadrants.

Thus, this affirmative is sometimes true. Every point but (0, 0) that have same x and y values are in the 1st or 3rd quadrant except for (0, 0).