Find two positive numbers x and y such that t | vedclass

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(A) Let one number be $x$. Then,the other number is $y = 35 - x$.Let $P(x) = x^{2} y^{5}$. Substituting $y$,we get:$P(x) = x^{2} (35 - x)^{5}$.To find

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$x=10, y=25$

Vedclass · Answer

(A) Let one number be $x$. Then,the other number is $y = 35 - x$.Let $P(x) = x^{2} y^{5}$. Substituting $y$,we get:$P(x) = x^{2} (35 - x)^{5}$.To find the critical points,we differentiate $P(x)$ with respect to $x$:$P'(x) = 2x(35 - x)^{5} + x^{2} \cdot 5(35 - x)^{4} \cdot (-1)$$P'(x) = x(35 - x)^{4} [2(35 - x) - 5x]$$P'(x) = x(35 - x)^{4} (70 - 2x - 5x)$$P'(x) = x(35 - x)^{4} (70 - 7x) = 7x(35 - x)^{4} (10 - x)$.Setting $P'(x) = 0$,we get $x = 0$,$x = 35$,or $x = 10$.Since $x$ and $y$ are positive numbers,$x = 0$ and $x = 35$ are rejected as they lead to a product of $0$.Now,we check the second derivative $P''(x)$ at $x = 10$:$P''(x) = \frac{d}{dx} [7x(35 - x)^{4} (10 - x)]$.Using the product rule,at $x = 10$,the term $(10 - x)$ becomes $0$,so:$P''(10) = 7(10)(35 - 10)^{4} (-1) = -70(25)^{4} Since $P''(10) Thus,$x = 10$ and $y = 35 - 10 = 25$.

Find two positive numbers $x$ and $y$ such that their sum is $35$ and the product $x^{2} y^{5}$ is a maximum.

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