01 - 04 Number Problems in Maxima and Minima

Let
x = the number and
x² = the square of the number
y = the difference between x and x²
 

$y = x - x^2$

$y' = 1 - 2x = 0$

$x = \frac{1}{2}$           answer

Let
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

Let
x and y = the numbers
z = sum of their squares
 

$k = x + y$

$y = k - x$

 
$z = x^2 + y^2$

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

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

$2x - k = 0$

$x = \frac{1}{2}k$
 

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

$y = \frac{1}{2}k$
 

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

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

Let
x and y = the numbers
z = sum of their cubes
 

$k = x + y$

$y = k - x$
 

$z = x^3 + y^3$

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

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

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

$x = \frac{1}{2}k$
 

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

$y = \frac{1}{2}k$
 

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

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