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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x+y=60$,so $y=60-x$. Let $f(x) = x y^3 = x(60-x)^3$. To find the maximum,we find the derivative $f'(x)$: $f'(x) = (60-x)^3 + x \cdot 3(60-x

Vedclass · Accepted Answer

$x=15, y=45$

Vedclass · Answer

(A) Given $x+y=60$,so $y=60-x$. Let $f(x) = x y^3 = x(60-x)^3$. To find the maximum,we find the derivative $f'(x)$: $f'(x) = (60-x)^3 + x \cdot 3(60-x)^2(-1) = (60-x)^2 [60-x - 3x] = (60-x)^2 (60-4x)$. Setting $f'(x) = 0$,we get $x=60$ or $x=15$. Since $x$ and $y$ are positive numbers and $x+y=60$,$x$ cannot be $60$ (as that would make $y=0$). Thus,$x=15$. Now,$y = 60 - 15 = 45$. Checking the second derivative $f''(x) = 2(60-x)(-1)(60-4x) + (60-x)^2(-4) = -2(60-x)(60-4x + 2(60-x)) = -2(60-x)(180-6x)$. For $x=15$,$f''(15) = -2(45)(180-90) = -2(45)(90) Since the second derivative is negative,$x=15$ is a point of local maxima. Therefore,the numbers are $x=15$ and $y=45$.

Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum.

Similar Questions

The ordinate of all points on the curve $y = \frac{1}{2\sin^2 x + 3\cos^2 x}$ where the tangent is horizontal,is

The least area of a circle circumscribing any right triangle of area $S$ is

$A$ wire of length $36 \text{ m}$ is cut into two pieces. One piece is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum,and the circumference of the circle is $k \text{ m}$,then $\left(\frac{4}{\pi}+1\right) k$ is equal to ..... .

The maximum value of ${\left( {\frac{1}{x}} \right)^x}$ is

Two positive numbers $x$ and $y$ are such that $(x+y)=60$ and $x y^3$ is maximum. Then the numbers $x$ and $y$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module