05 - 08 Number Problems in Maxima and Minima
A D V E R T I S E M E N T
Problem 5
The sum of two positive numbers is 2. Find the smallest value possible for the sum of the cube of one number and the square of the other.
Solution 5
Let x and y = the numbers
x + y = 2 Equation (1)
1 + y’ = 0
y’ = –1
z = x3 + y2 Equation (2)
dz/dx = 3x2 + 2y y’ = 0
3x2 + 2y(–1) = 0
y = (3/2) x2
From Equation (1)
x + (3/2) x2 = 2
2x + 3x2 = 4
3x2 + 2x – 4 = 0
x = 0.8685 & –1.5352
use x = 0.8685
y = (3/2)(0.8685)2
y = 1.1315
z = 0.86853 + 1.13152
z = 1.9354 answer
Problem 6
Find two numbers whose sum is a, if the product of one to the square of the other is to be a minimum.
Solution:
Let x and y = the numbers
x + y = a
x = a – y
z = xy2
z = (a – y) y2
z = ay2 – y3
dz/dy = 2ay – 3y2 = 0
y = 2/3 a
x = a – 2/3 a
x = 1/3 a
The numbers are 1/3 a, and 2/3 a. answer
Problem 7
Find two numbers whose sum is a, if the product of one by the cube of the other is to be a maximum.
Solution:
Let x and y the numbers
x + y = a
x = a – y
z = xy3
z = (a – y) y3
z = ay3 – y4
dz/dy = 3ay2 – 4y3 = 0
y2 (3a – 4y) = 0
y = 0 (absurd) and 3/4 a (use)
x = a - 3/4 a
x = 1/4 a
The numbers are 1/4 a and 3/4 a. answer
Problem 8
Find two numbers whose sum is a, if the product of the square of one by the cube of the other is to be a maximum.
Solution:
Let x and y the numbers
x + y = a
1 + y’ = 0
y = –1
z = x2 y3
dz/dx = x2 (3y2 y’) + 2xy3 = 0
3x(–1) + 2y = 0
x = 2/3 y
2/3 y + y = a
5/3 y = a
y = 3/5 a
x = 2/3 (3/5 a)
x = 2/5 a
The numbers are 2/5 a and 3/5 a. answer
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