find two numbers whose sum is 15 and the square of one multiplied the cube of the other is maximum
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15 and 1 is that two number.
(15^3)(1^2)=3375
(15^3)(1^2)=3375
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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.
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.
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