# Calculus

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The product of 2 positive numbers is 48. find the value of the numbers if the sum of one of the numbers and the cube of the other is a minimum.

• Calculus -

a = first number

b = second number

a * b = 48 Divide both sides by a

b = 48 / a

S = the sum of one of the numbers and the cube of the other number

S = a + b ^ 3

S = a + ( 48 / a ) ^ 3

S = a + 110,592 / a ^ 3

S = a + 110,592 * a ^ - 3

First derivaton :

d S / d a = 1 - 3 * 110,592 * a ^ - 4

d S / d a = 1 - 331,776 / a ^ 4

Second derivation :

d ^ 2 S / d a ^ 2 = - 3 * 331,776 ( - 4 ) * a ^ - 5

d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5

A function has minimum or maximum in poit where first derivation = 0

If second derivaton < 0 function has maximum.

If second derivaton > 0 function has minimum.

In this case:

d S / d a = 1 - 331,776 / a ^ 4 = 0

1 = 331,776 / a ^ 4 Multiply both sides by a ^ 4

a ^ 4 = 331,776

a = fourth root of 331,776

a = + OR - 24

For a = - 24

d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5 =

1,327,104 / - 7,962,624 = - -0.166667 < 0

function has maxsimum.

For a = 24

d ^ 2 S / d a ^ 2 = 1,327,104 / a ^ 5 =

1,327,104 / -7,962,624 = 0.166667 > 0

function has minimum.

So a = 24

b = 48 / a = 48 / 24 = 2

The mumbers are a = 24 and b = 2

Local minimum = a + b ^ 3 = 24 + 2 ^ 3 = 24 + 8 = 32

• Calculus -

Thank you!!!

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