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Wednesday, September 11, 2013

Maximum Minimum Problem, 7

Category: Differential Calculus, Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the area of the largest rectangle having one side on the x-axis and inscribed in the triangle formed by the lines y = x , y = 0, and 3x + y = 20.

Solution:

To illustrate the problem, it is better to draw the figure and sketch the three given lines in Rectangular Coordinate System as follows

Photo by Math Principles in Everyday Life

The first thing that we have to do is to find the length of a rectangle which is x and then find the width of a rectangle which is y. 

To get the value of x for a rectangle, we need to use the lines y = x and 3x + y = 20. 

The line 3x + y = 20 can be written as







Hence, the length of a rectangle is









To get the value of y for a rectangle, we need to use the lines y = x and y = 0.

Hence, the width of a rectangle is 







We know that the area of a rectangle is







Next, we need to eliminate x at the above equation. We can use one of the three lines to substitute the value of x. The best is y = x because it is the simplest equation of the three lines. Substitute x = y to the above equation, we have





Take the derivative on both sides of the equation with respect to y, we have





Set dA/dy = 0 because we want to maximize the area of a rectangle.









Therefore,





Finally, the area of a rectangle is











Tuesday, September 10, 2013

Right Circular Cone - Sphere

Category: Differential Calculus, Solid Geometry, Algebra

"Published in Newark, California, USA"

Find the radius r of the right circular cone of maximum volume which can be inscribed in a sphere of radius R.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

We know that the volume of a right circular cone is 



Next, we need another equation in order to eliminate h at the above equation. Apply Pythagorean Theorem at the right triangle inside the right circular cone, we have






Substitute the value of h to the first equation, we have



Take the derivative on both sides of the equation with respect to r. Consider R as a constant because a right circular cone is inscribed in a sphere.






 







Equate dV/dr = 0 because we want to maximize the volume of a right circular cone



Divide both sides of the equation by ⅓ Ï€r, we have











Divide both sides of the equation by 2R, we have



Square on both sides of the equation to remove the radical sign, we have













Therefore,