# 01 - 04 Number Problems in Maxima and Minima

**Problem 1**

What number exceeds its square by the maximum amount?

**Solution 1**

x = the number and

x

^{2}= the square of the number

y = the difference between x and x

^{2}

$y = x - x^2$

$y' = 1 - 2x = 0$

$x = \frac{1}{2}$ *answer*

**Problem 2**

What positive number added to its reciprocal gives the minimum sum?

**Solution 2**

x = the required positive number and

1/x = the reciprocal of the number

y = sum of x and 1/x

$y = x + 1/x$

$y = x + x^{-1}$

$y' = 1 - x^{-2} = 0$

$x = 1$ *answer*

**Problem 3**

The sum of two numbers is k. Find the minimum value of the sum of their squares.

**Solution 3**

x and y = the numbers

z = sum of their squares

$k = x + y$

$y = k - x$

$z = x^2 + y^2$

$z = x^2 + (k - x)^2$

$dz/dx = 2x + 2(k - x)(-1) = 0$

$2x - k = 0$

$x = \frac{1}{2}k$

$y = k - \frac{1}{2}k$

$y = \frac{1}{2}k$

$z = (\frac{1}{2}k)^2 + (\frac{1}{2}k)^2$

$z = \frac{1}{2}k^2$

**Problem 4**

The sum of two numbers is k. Find the minimum value of the sum of their cubes.

**Solution 4**

x and y = the numbers

z = sum of their cubes

$k = x + y$

$y = k - x$

$z = x^3 + y^3$

$z = x^2 + (k - x)^3$

$dz/dx = 3x^2 + 3(k - x)^2(-1) = 0$

$x^2 - (k^2 - 2kx + x^2) = 0$

$x = \frac{1}{2}k$

$y = k - \frac{1}{2}k$

$y = \frac{1}{2}k$

$z = (\frac{1}{2}k)^3 + (\frac{1}{2}k)^3$

$z = \frac{1}{4}k^3$