Rectangular Parallelepiped Problem

Category: Solid Geometry

"Published in Suisun City, California, USA"

A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side x at each corner and then folding up the sides. Express the volume V of the box as a function of x. 

Solution:

To visualize the problem, let's draw the figure first as follows:

Photo by Math Principles in Everyday Life

If you cut and remove the four squares of side x and then fold up the four sides, the figure is a rectangular parallelepiped. 

Photo by Math Principles in Everyday Life

The volume of a rectangular parallelepipied is given by the formula

                             V = L W H

Substitute the given dimensions, we have

                             V = (20 - 2x)(12 - 2x)(x)

                             V = [2(10 - x)][2(6 - x)](x)

                             V = 4x (10 - x)(6 - x)

Therefore,

                         V(x) = 4x (10 - x)(6 - x)

                         where  0 < x < 6 
      

Maximum Volume - Right Circular Cylinder

Category: Differential Calculus, Solid Geometry

"Published in Suisun City, California, USA"

A right circular cylinder is inscribed in a sphere with radius R. Find the largest possible volume of such a cylinder. 

Solution:

To visualize the problem, let's draw the figure first. Inscribed means inside and so a right circular cylinder is located inside the sphere. 

Photo by Math Principles in Everyday Life

Next, we have to find the dimensions of a right circular cylinder in order to get its volume. By labeling the figure further, we have

Photo by Math Principles in Everyday Life

By Pythagorean Theorem,

We know that the volume of a right circular cylinder is

but h = 2a

Equate r² on both sides of the equation,

Take the derivative of both sides of the equation with respect to a and equate it to zero because we want to maximize the volume of a right circular cylinder. Consider R as a constant in the equation.

Now, we can solve for r,

Therefore, 

Two Parallel Lines

Category: Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the points of intersection of the following lines:

                        x + 2y = 6

                      2x + 4y = -9

Solution: 

Since the given equations are all first degree, then they are linear equations. They are straight lines. We can graph the two lines by getting their slope and y-intercept. 

For x + 2y = 6, 

                        x + 2y = 6

                              2y = - x + 6

                                y = -½ x + 3

                        slope (Δy/Δx), m = -½

                        y-intercept, b = 3

To trace the graph, plot 3 at the y-axis. This is your first point of the line (0, 3). Next, use the slope to get the second point. From the first point, count 2 units to the left and then 1 unit upward. 

For 2x + 4y = -9,

                        2x + 4y = -9

                               4y = - 2x - 9

                                 y = -½ x - ⁹/₄

                        slope (Δy/Δx), m = -½  

                        y-intercept, b = - ⁹/₄

To trace the graph, plot - ⁹/₄ at the y-axis. This is your first point of the line (0, - ⁹/₄). Next, use the slope to get the second point. From the first point, count 2 units to the left and then 1 unit upward.   

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From the graph, the two lines are parallel because their slopes are the same which is -½. The two lines will never meet how far they are extended. When you solve for x and y from the two given equations, their x and y will be equal to zero. From the two given equations,

                        x + 2y = 6

                      2x + 4y = -9

Multiply the first equation by 2 and -1 at the second equation. Add the two equations and let's see what will happen to x and y.

            2 (x + 2y = 6)                 2x + 4y = 12
                                       →
        - 1 (2x + 4y = -9)                -2x - 4y = 9
                                              ______________

                                                           0 ≠ 21

Since their x and y are equal to zero, then we cannot solve for x and y. Also, the right side of the final equation is not zero. Therefore, the two lines are parallel to each other.