Need some help on this calculus 3 question.

Question

A rectangular box is placed in the first octant of x, y, z space so that one corner is at the origin. The opposite corner lies on the surface x^2+y^2+z = 1. Find the point which gives the maximum possible volume.

Answer 1

Given:

Cube Volume: V = xyz

One corner of the cube lies on the surface defined by: x² + y² + z = 1

Find:

( x, y, z ) = (?, ?, ?)

Assume:

x > 0

y > 0

z > 0

For the volume to be a real-world volume.

Solution:

Step 1: Rearrange the equation for the surface for z in terms of x's and y's

z = 1 - x² - y²

Step 2: Substitute the surface equation for z into the volume equation

V = xy( 1 - x² - y² )

Step 3: Take the partial derivative in respect to x, set it equal to 0 (this will give either local mins or maxs)

∂V/∂x = V_x = xy * ( -2 x ) + y ( 1 - x² - y² ) = y - 3x²y - y³ = y ( 1 - 3x² - y² ) = 0

This means either:

y = 0 OR 1 - 3x² - y² = 0 → y² = 1 - 3x² (I'm strategically not going to take the square root, tempting as it may be)

The y = 0 solution cannot happen due to the assumption that x, y, and z cannot be zero nor negative.

Step 4: Take the partial derivative in respect to y, set it equal to 0

V_y = xy * ( - 2 y ) + x ( 1 - x² - y² ) = x - 3xy² - x³ = x ( 1 - 3y² - x² ) = 0

This means either:

x = 0 OR 1 - 3y² - x² = 0 → x² = 1 - 3y² (do you see why I didn't square root it? it fits perfectly right into that y² spot in this equation)

The x = 0 solution cannot happen due to the assumption that x, y, and z cannot be zero nor negative.

Step 5: Plug in the equation for y² into this equation for x² and solve for x

x² = 1 - 3 ( 1 - 3x² ) = 1 - 3 + 9x² → 8x² = 2 → x² = 1/4 For x > 0, x ≠ - 1/2

∴ x = 1/2

Step 6: Use this newly found solution for x to solve for y

y² = 1 - 3 ( 1/2 )² = 1 - 3/4 = 1/4 For y > , y ≠ -1/2

∴ y = 1/2

Back to where this problem all started off

Step 7: Plug the solutions for x and y to solve for z

z = 1 - 1/4 - 1/4

∴ z = 1/2

Step 8: Calculate this maximum volume

∴ V = 1/2 * 1/2 * 1/2 ∴ V = 1/8 = 0.125 un³

I hope this helps! Message me in the comments with any questions, comments, or concerns you have with the handling of the problem above!

It should be noted that we excluded the x = 0 and y = 0 "solutions" because they didn't apply to real-world applications, but it was a local extreme of the function, just not the one we were looking for; it is a local minimum for the volume.