How do you find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes, and one vertex in the plane x+7y+11z=77?

Question

Ezra Adams · Accepted Answer

the volume is # 5929/27 approx 219.6 " cubic units"#

you're optimising volume function

#f(x,y,z) = xyz#

with constraint #g(x,y,z) = x + 7y + 11z = 77#

using the Lagrange Multiplier ie #nabla f = lambda nabla g#

we get

#[(yz),(xz),(xy)] = lambda [(1),(7),(11)]#

comparing #lambda#'s and from rows 1 and 2

#(yz)/1 = (xz)/7 implies x = 7y#

from rows 2 and 3

#(xz)/7 = (xy)/11 implies z = (7y)/11#

if we park those back into the constraint

#7y + 7y + 11( (7y)/11 ) = 77#

#implies y = 77/21#

so #x = 77/3, z = 7/3#

So the volume is #77/3 * 77/21 * 7/3 = 5929/27 approx 219.6 " cubic units"#

in terms of scoping for a reality check, the total volume under that plane in the first octant is

#\int_(x=0)^(77) quad \int_(y=0)^( 11 - x/7) quad (77-x-7y)/11 quad dy \ dx approx 988 " cubic units"#

Isabella Ackerman · Answer

undefined To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane $x + 7y + 11z = 77$, follow these steps:

1. Set up the problem by defining variables for the dimensions of the box.
2. Write an expression for the volume of the box in terms of these variables.
3. Express one of the dimensions in terms of the other dimensions using the given constraints.
4. Use calculus to find the maximum volume of the box.

Let $ x $, $ y $, and $ z $ represent the dimensions of the box. Then, the volume ($ V $) of the box is given by $ V = xyz $.

The constraint given by the plane equation $ x + 7y + 11z = 77 $ can be rearranged to express $ z $ in terms of $ x $ and $ y $ as follows:
$$ z = \frac{77 - x - 7y}{11} $$

Now, substitute this expression for $ z $ into the volume equation:
$$ V = x \cdot y \cdot \left(\frac{77 - x - 7y}{11}\right) $$

Next, find the critical points of $ V $ by taking partial derivatives with respect to $ x $ and $ y $, setting them equal to zero, and solving for $ x $ and $ y $.

After finding the critical points, determine which critical point corresponds to the maximum volume. This can be done by using the second derivative test or by evaluating the volume at each critical point.

Once you find the dimensions $ x $, $ y $, and $ z $ corresponding to the maximum volume, substitute them into the volume equation to find the maximum volume of the rectangular box.

In summary:
1. Define variables for the dimensions of the box.
2. Write the volume expression in terms of these variables.
3. Express one dimension in terms of the others using the given constraint.
4. Find critical points of the volume function.
5. Determine the maximum volume among the critical points.
6. Substitute the dimensions of the maximum volume into the volume expression to find the maximum volume of the rectangular box.