# A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant of the paraboloid #z = 4-x^2-y^2#, what is the box's maximum volume?

2 cubic units

we can do this 2 ways, first just by re-writing the volume in 2 variables as

we can then look for critical points using the partial derivatives

The trivial solution is at (0,0) when the volume is zero. We can ignore that for physical reasons.

combining these we have

but for now we can do this a second way, by using the method of Lagrange Multipliers.

We have function

Same as before.

graph{4-x^2 [-10, 10, -5, 5]}

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To find the maximum volume of the rectangular box, we need to maximize the product of its three dimensions: length, width, and height.

Let's denote the dimensions of the box as ( x ), ( y ), and ( z ). Since the box has three of its faces on the coordinate planes, its vertices are at ( (0,0,0) ), ( (x,0,0) ), ( (0,y,0) ), and ( (0,0,z) ).

The volume of the box is given by ( V = xyz ).

Given that one vertex of the box lies in the first octant of the paraboloid ( z = 4 - x^2 - y^2 ), we have the constraint that ( z = 4 - x^2 - y^2 ).

We want to maximize ( V = xyz ) subject to the constraint ( z = 4 - x^2 - y^2 ).

Using the constraint, we can express ( z ) in terms of ( x ) and ( y ) as ( z = 4 - x^2 - y^2 ).

Now, we have ( V = xy(4 - x^2 - y^2) ).

To maximize ( V ), we take the partial derivatives of ( V ) with respect to ( x ) and ( y ), set them equal to zero, and solve for ( x ) and ( y ).

Taking partial derivatives:

( \frac{\partial V}{\partial x} = y(4 - 3x^2 - y^2) )

( \frac{\partial V}{\partial y} = x(4 - x^2 - 3y^2) )

Setting them equal to zero:

( y(4 - 3x^2 - y^2) = 0 )

( x(4 - x^2 - 3y^2) = 0 )

The solutions to these equations occur when either ( x = 0 ), ( y = 0 ), or both terms in parentheses equal zero.

Solving these equations, we find the critical points to be ( (0, 0) ), ( (\pm \sqrt{3}, 0) ), and ( (0, \pm \sqrt{3}) ).

To find which of these points gives the maximum volume, we can use the second partial derivative test or evaluate the volume at each critical point.

However, we notice that the function ( V = xy(4 - x^2 - y^2) ) is symmetric about the ( x )-axis and ( y )-axis, so we can restrict our search to the first quadrant.

Evaluating the volume at ( (0, 0) ), ( (\sqrt{3}, 0) ), and ( (0, \sqrt{3}) ), we find that the maximum volume occurs at ( (\sqrt{3}, \sqrt{3}) ) and its symmetric points.

Therefore, the maximum volume of the rectangular box is achieved when ( x = y = \sqrt{3} ), and ( z = 4 - (\sqrt{3})^2 - (\sqrt{3})^2 = 4 - 3 = 1 ).

Thus, the maximum volume of the box is ( V = \sqrt{3} \times \sqrt{3} \times 1 = 3 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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