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

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. 

option a is correct.

only one solution to the system of equations

step-by-step explanation:

if the graph of the equations do not intersects(i.,e if they are parallel) then, there are no solutions that are true for both the equations.

if the graph of the equations intersects, then there is one solution and that is true for both the equations.

since, in the given figure:

the two lines intersect each other

⇒there is only one solution which satisfy both the lines of an equations.

therefore, there is only one solution to the system of equation