Calculus

Application of Derivatives

Medium

core

The sum of two positive numbers is 60. If the sum of their squares in minimum, then the absolute value of the difference of their cubes is

56000

11500

87750

✅ Correct Option: 1

Let the two positive numbers be $x$ and $y$ .

Given: $x + y = 60$

Therefore: $y = 60 - x$

To minimize the sum of squares $x^2 + y^2$ , substitute $y = 60 - x$ :

$x^2 + (60 - x)^2$

$= x^2 + 3600 - 120x + x^2$

$= 2x^2 - 120x + 3600$

Taking the derivative with respect to $x$ :

$\frac{d}{dx}(2x^2 - 120x + 3600) = 4x - 120$

Setting equal to zero:

$4x - 120 = 0$

$4x = 120$

$x = 30$

Since $x = 30$ :

$y = 60 - 30 = 30$

The absolute value of the difference of their cubes:

$|x^3 - y^3|$

$= |30^3 - 30^3|$

$= |27000 - 27000|$

$= 0$

Therefore, the absolute value of the difference of their cubes is $0$ .

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B. Decreasing in $(-2,3)$

C. Neither increasing nor decreasing in $(-2,3)$

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Choose the correct answer from the options given below:

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Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$y = x^2$	I.	decreasing on $(0, \infty)$
B.	$y = -x^2$	II.	increasing on $(0, \infty)$
C.	$y = x$	III.	decreasing on $(-\infty, \infty)$
D.	$y = -x$	IV.	increasing on $(-\infty, \infty)$

Choose the correct answer from the options given below:

The sum of two positive numbers is 60. If the sum of their squares in minimum, then the absolute value of the difference of their cubes is

Solution

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