Question

Find two numbers whose sum is 15 and when the sqaure of one multiplied by the cube of other is maximum

Answer 1

Step-by-step explanation:

Lets consider the two numbers to be x,y (both greater than 0).

So x+y = 15

Now to maximise T = x^2 * y^3

So T = y^3 (15-y)^2 = y^3 { 225 + y^2 - 30y)

So T = 225y^3 + y^5 - 30y^4.

Now maximise this using dT/dy = 0.

Clearly y=0 will not maximise "T".

Hence y(2y-30) + 3 { 225 + y^2 - 30y } = 0.

Solve this QE, for y value (for which the maximum occurs).

Answer 2

Answer:

First form a equations i. x+y=15 ----(1) and (x^2)(y^3) is maximum

Now take the value of x from equation (1).

=((15-y) 2 )(y 3 )

=(225+y 2 -30y)(y 3 )

=(225y 3 +y 5 -30y 4 )

Now differentiate in terms of y and putting it equal to 0 which is the slope of the graph at the maxima or minima

0=675y 2 +5y 4 -120y 3

0=5y 2 (y 2 -24y+135)

y2-24y+135=0

The above equation are satisfied only at 15,9

At 15,0 it is at the minima, So y=9 and x=6 must be the correct answers

Step-by-step explanation:

Lets consider the two numbers to be x,y (both greater than 0).

So x+y = 15

Now to maximise T = x^2 * y^3

So T = y^3 (15-y)^2 = y^3 { 225 + y^2 - 30y)

So T = 225y^3 + y^5 - 30y^4.

Now maximise this using dT/dy = 0.

Clearly y=0 will not maximise "T".

Hence y(2y-30) + 3 { 225 + y^2 - 30y } = 0.

Solve this QE, for y value (for which the maximum occurs).

Hope that helps.

Best Regards,

or

Let the two numbers be x and y.

Given that the sum of two numbers is 15.

Therefore,

x + y = 15  ........ (1)

We have to find the two numbers with the condition that the square of one number multiplied by the cube of the other is maximum.

Hence,

T = x² . y³  

T = y³ . (15 - y)²

T = y³ . (225 + y² - 30y)

dT / dy = y³ . (2y - 30) + 3 y² (225 + y² - 30y)

dT / dy = 2y⁴ - 30y³ + 675y² + 3y⁴ - 90y³

dT / dy = 5y⁴ - 120y³ + 675y²

dT / dy = 5y² (y² - 24y + 135)

5y² (y² - 24y + 135) = 0

5y² = 0  ,  y² - 24y + 135 = 0

y² = 0    ,  y² - 15y - 9y +135 = 0

y = 0     ,  (y² - 15y) - (9y -135) = 0

               y (y - 15) - 9 (y - 15) = 0

                 (y - 15) (y - 9) = 0

                 y = 15 , y = 9

Thus values of y are 0 , 9 , 15

Now,

for y = 0, x = 15 - 0 = 15

for y = 9, x = 15 - 9 = 6

for y = 15, x = 15 - 15 = 0

Thus values of x are 0 , 6 , 15

This is the required solution.