88. Find two positive numbers x and y such that their sum is 35 and the product x2 y5 is a maximum.

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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

88. Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum.

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Vishal Baghel | Contributor-Level 10

4 months ago

We have, x + y = 35.

y = 35 - x

Let the product, P =x²y⁵

$\Rightarrow$ P = x² (35 -x)⁵

So, $\frac{d p}{d x}$ = x² 5 (35 -x)⁴ (1) + (35 -x)⁵ 2x

= x (35 -x)⁴ [ - 5x + (35 -x) 2]

= x (35 -x)⁴ [ - 5x + 70 - 2x]

= x (35 -x)⁴ (70 - 7x)

= 7x (35 -x)⁴ (10 -x)

At $\frac{d p}{d x} = 0$

$\Rightarrow$ 7x (35 -x)⁴ (10 -x) = 0

$\Rightarrow$ x = 0, 35, 10

As x is a (+) ve number we have only

x = 10, 35

And again (at x = 35) y = 35 = 0 but yis also a (+) ve number

we get, x = 10 (only)

whenx < 10,

$? \frac{d p}{d x} = (+ v e) (+ v e) (+ v e) > 0$

and when x > 10,

$\frac{d p}{d x} = (+ v e) v e (+ v e) (- v e) = (- v e) < 0$

$∴ \frac{d p}{d x}$ changes from (+ ve) to ( -ve) as x increases while passing through 10

Hence, x = 10 is a point of local maxima

So, y = 35 - 10 = 25

∴x = 10 and y = 25

We have, x + y = 35.y = 35 - xLet the product, P =x2 y5 <math><mrow><mo>⇒</mo></mrow></math> P = x2 (35 -x)5So,  <math><mrow><mfrac><mrow><mi>d</mi><mi>p</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac></mrow></math> = x2 5 (35 -x)4 (1) + (35 -x)5 2x= x (35 -x)4 [ - 5x + (35 -x) 2]= x (35 -x)4 [ - 5x + 70 - 2x]= x (35 -x)4 (70 - 7x)= 7x (35 -x)4 (10 -x)At <math><mrow><mfrac><mrow><mi>d</mi><mi>p</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>0</mn></mrow></math><math><mrow><mo>⇒</mo></mrow></math>7x (35 -x)4 (10 -x) = 0<math><mrow><mo>⇒</mo></mrow></math> x = 0, 35, 10As x is a (+) ve number we have onlyx = 10, 35And again (at x = 35) y = 35 = 0 but yis also a (+) ve numberwe get, x = 10 (only)whenx < 10, <math><mrow><mo>? </mo><mfrac><mrow><mi>d</mi><mi>p</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mo stretchy="false"> (</mo><mo>+</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo stretchy="false"> (</mo><mo>+</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo stretchy="false"> (</mo><mo>+</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo>></mo><mn>0</mn></mrow></math>and when x > 10, <math><mrow><mfrac><mrow><mi>d</mi><mi>p</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mo stretchy="false"> (</mo><mo>+</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mi>v</mi><mi>e</mi><mo stretchy="false"> (</mo><mo>+</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo stretchy="false"> (</mo><mo>−</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo>=</mo><mo stretchy="false"> (</mo><mo>−</mo><mi>v</mi><mi>e</mi><mo stretchy="false">)</mo><mo><</mo><mn>0</mn></mrow></math><math><mrow><mo>∴</mo><mfrac><mrow><mi>d</mi><mi>p</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac></mrow></math> changes from (+ ve) to ( -ve) as x increases while passing through 10Hence, x = 10 is a point of local maximaSo, y = 35 - 10 = 25∴x = 10 and y = 25

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