The-sum-of-two-positive-numbers-is-20-find-the-numbers-i-If-their-product-is-maximum-ii-If-the-sum-of-their-square-is-maximum-iii-If-the-product-of-the-square-of-one-and-the-cube-of-the-othe

Question Number 12796 by tawa last updated on 01/May/17

The sum of two positive numbers is 20 . find the numbers

(i) If their product is maximum

(ii) If the sum of their square is maximum

(iii) If the product of the square of one and the cube of the other is maximum

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

i) a + b = 20, a b = m a x \Rightarrow a = b = 10

o r :

f (a) = a (20 - a) = 20 a - a^{2}

d f / d a = 20 - 2 a = 0 \Rightarrow a = 10 = b

i i) f (a) = a^{2} + {(20 - a)}^{2} = 2 a^{2} - 40 a + 400

d f / d a = 4 a - 40 = 0 \Rightarrow a = 10 = b .

i i i) f (a) = a^{2} + {(20 - a)}^{3} o r a^{3} + {(20 - a)}^{2}

\Rightarrow d f / d a = [2 a - 3 {(20 - a)}^{2}] o r [3 a^{2} - 2 (20 - a)]

1) 2 a - 3 (400 - 40 a + a^{2}) = 0

3 a^{2} - 122 a + 1200 = 0

a = \frac{122 \pm \sqrt{122^{2} - 4 \times 3 \times 1200}}{6} = \frac{122 \pm 22}{6} = 22, 16.6

a = 16.6, b = 3.4

2) 3 a^{2} - 2 (20 - a) = 0 \Rightarrow 3 a^{2} + 2 a - 40 = 0

a = \frac{- 2 \pm \sqrt{4 + 3 \times 4 \times 40}}{6} = \frac{- 2 \pm 22}{6} = - 4, 3.3

a = 3.3, b = 16.7

Commented by tawa last updated on 01/May/17

wow, God bless you sir .

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