# 05 - 08 Number Problems in Maxima and Minima

**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**

$x + y = 2$ → Equation (1)

$1 + y' = 0$

$y' = -1$

$z = x^3 + y^2$ → Equation (2)

$dz/dx = 3x^2 + 2y \, y' = 0$

$3x^2 + 2y(-1) = 0$

$y = \frac{3}{2}x^2$

From Equation (1)

$x + \frac{3}{2}x^2 = 2$

$2x + 3x^2 = 4$

$3x^2 + 2x - 4 = 0$

$x = 0.8685 \, \text{&} \, -1.5352$

Use

$x = 0.8685$

$y = \frac{3}{2}(0.8685^2)$

$y = 1.1315$

$z = 0.8685^3 + 1.1315^2$

$z = 1.9354$ *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 maximum.

**Solution 6**

$x + y = a$

$x = a - y$

$z = xy^2$

$z = (a - y)y^2$

$z = ay^2 - y^3$

$dz/dy = 2ay - 3y^2 = 0$

$y = \frac{2}{3}a$

$x = a - \frac{2}{3}a$

$x = \frac{1}{3}a$

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 7**

$x + y = a$

$x = a - y$

$z = xy^3$

$z = (a - y)y^3$

$z = ay^3 - y^4$

$dz/dy = 3ay^2 - 4y^3 = 0$

$y^2(3a - 4y) = 0$

$y = 0 \, \text{(absurd) and} \, \frac{3}{4}a \, \text{(use)}$

$x = a - \frac{3}{4}a$

$x = \frac{1}{4}a$

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 8**

$x + y = a$

$1 + y' = 0$

$y = -1$

$z = x^2 y^3$

$dz/dx = x^2(3y^2 \, y') + y^3(2x) = 0$

$3x \, y' + 2y = 0$

$3x(-1) + 2y = 0$

$x = \frac{2}{3}y$

$\frac{2}{3}y + y = a$

$\frac{5}{3}y = a$

$y = \frac{3}{5}a$

$x = \frac{2}{3}(\frac{3}{5}a)$

$x = \frac{2}{5}a$

The numbers are 2/5 a and 3/5 a. *answer*