Divide 15 into two parts such that the square of one multiplied with t

1.	Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.
Answer» Let the two numbers be a and b. Given : a + b = 15 Assume, S = a²b³ S = a²(15 - a)³ \(\frac{dS}{da}\)= 2a(15 - a)³ - 3a²(15 - a)² Condition maxima and minima is, ⇒\(\frac{dS}{da}\) = 0 ⇒ 2a(15 - a)³ - 3a²(15 - a)² = 0 ⇒ a(15 - a)² {2(15 – a) – 3a} = 0 ⇒ a(15 - a)² {30 – 5a} = 0 ⇒ a = 0, 15, 6 \(\frac{d^2S}{da^2}\) = 2(15 - a)³ - 6a(15 - a)² - 6a(15 - a)²+ 3a²(15 - a) ⇒ \(\frac{d^2S}{da^2}\) = (15 - a){2(15 - a)² - 12a(15 - a)+ 3a²} For S to minimum, \(\frac{d^2S}{da^2}\) > 0 a = 0 ⇒ \(\frac{d^2S}{da^2}\) = 450 ⇒ minima a = 6 ⇒ \(\frac{d^2S}{da^2}\) = - 378 ⇒ maxima a = 15 ⇒ \(\frac{d^2S}{da^2}\) = 0 ⇒ point of inflection Hence, two numbers are 0 and 15

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Answer»

Let the two numbers be a and b.

Given :

a + b = 15

Assume,

S = a²b³

S = a²(15 - a)³

\(\frac{dS}{da}\)= 2a(15 - a)³ - 3a²(15 - a)²

Condition maxima and minima is,

⇒\(\frac{dS}{da}\) = 0

⇒ 2a(15 - a)³ - 3a²(15 - a)² = 0

⇒ a(15 - a)² {2(15 – a) – 3a} = 0

⇒ a(15 - a)² {30 – 5a} = 0

⇒ a = 0, 15, 6

\(\frac{d^2S}{da^2}\) = 2(15 - a)³ - 6a(15 - a)² - 6a(15 - a)²+ 3a²(15 - a)

⇒ \(\frac{d^2S}{da^2}\) = (15 - a){2(15 - a)² - 12a(15 - a)+ 3a²}

For S to minimum,

\(\frac{d^2S}{da^2}\) > 0

a = 0 ⇒ \(\frac{d^2S}{da^2}\) = 450 ⇒ minima

a = 6 ⇒ \(\frac{d^2S}{da^2}\) = - 378 ⇒ maxima

a = 15 ⇒ \(\frac{d^2S}{da^2}\) = 0 ⇒ point of inflection

Hence, two numbers are 0 and 15