Applied Mathematics

Calculus

Medium

applied

Two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum are :

$x = 30, y = 30$

$x = 20, y = 40$

$x = 15, y = 45$

$x = 55, y = 5$

✅ Correct Option: 3

Substitute $y = 60 - x$ , so $f(x) = x(60-x)^3$ . Then $f'(x) = (60-x)^3 - 3x(60-x)^2 = (60-x)^2(60 - 4x)$ . Setting $f'(x) = 0$ in the interior gives $x = 15$ , so $y = 45$ .

2023: 7 June Shift 1

If the cost function and the profit function for a company is given by $C = 10 - 0.3x^2$ and $P = 0.3x^2 + 2x - 10$ respectively , where X represent units of output, then the revenue function is given by :

2023: 7 June Shift 1

The minimum value of $f(x) = 4x^3 - 48x + 105$ in the interval [1,3] is :

2026: 14 May Shift 2

$\int_1^4 x^3\log x\, dx$ is equal to:

2026: 21 May Shift 2

The demand function for a commodity is $p = 45 + x - x^2$ , then the consumer's surplus at equilibrium price $p_0 = 25$ will be:

2023: 7 June Shift 1

Match List - I with list- II

List-I , Differential equation	List - II , Degree
A. $\left(\frac{dy}{dx}\right)^3 + yx = 0$	2
B. $e^{\frac{dy}{dx}} + y^2 + y'' = 0$	1
C. $Xyy'' + x(y')^2 - yy' = 0$	Not defined
D. $(Y'')^2 + y = 0$	3

Choose the correct option below :

2023: 22 May Shift 3

Match List - I with List - II.

	List - I		List - II
(A)	If $y = x^2 - 8$ and $\frac{dy}{dx} = 0$ , then $x = ?$	(I)	1
(B)	If $p(x) = 3x + 1$ , then $R(x)$ at $x = 2$	(II)	0
(C)	If $y = x^3$ , then $\frac{dy}{dx}$ at $x = -1$	(III)	14
(D)	If $C(x) = 100 + 5x$ , $R(x) = 102 + 3x$ , then break-even point	(IV)	3

Choose the correct answer from the options given below :

Two positive numbers xxx and yyy such that x+y=60x + y = 60x+y=60 and xy3xy^3xy3 is maximum are :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum are :