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Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
- a)x = 45, y = 15
- b)x = 15, y = 45
- c)x = 10, y = 50
- d)x = 30, y = 30
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Find two positive numbers x and y such that x + y = 60 andxy3 is maxim...
two positive numbers x and y are such that x + y = 60.
x + y = 60
⇒ x = 60 – y ...(1)
Let P = xy3
∴ P =(60 – y)y3 = 60y3 – y4
Differentiating both sides with respect to y, we get
For maximum or minimum dP/dy = 0
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)
Thus, the two positive numbers are 15 and 45.
x + y = 60
⇒ x = 60 – y ...(1)
Let P = xy3
∴ P =(60 – y)y3 = 60y3 – y4
Differentiating both sides with respect to y, we get
For maximum or minimum dP/dy = 0
⇒ 180y2 - 4y3 = 0
⇒ 4y2 (45 - y) = 0
⇒ y = 0 or 45 - y = 0
⇒ y = 0 or y = 45
⇒ y = 45 (∵ y = 0 is not possible)
Thus, the two positive numbers are 15 and 45.
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Community Answer
Find two positive numbers x and y such that x + y = 60 andxy3 is maxim...
Explanation:
To find the values of x and y such that the product xy^3 is maximum, we need to maximize both x and y.
Given that x * y = 60, we can rewrite this equation as y = 60/x.
To maximize the product xy^3, we substitute the value of y from the above equation into the product:
P = x * (60/x)^3
Simplifying the equation, we have:
P = 60^3 / x^2
Maximizing x:
To maximize the product, we need to minimize the denominator, which means we need to maximize x.
To do this, we can take the derivative of P with respect to x and set it equal to zero to find the critical points:
dP/dx = -2 * 60^3 / x^3 = 0
Solving for x, we find x = 60.
Maximizing y:
Substituting the value of x = 60 into the equation y = 60/x, we find y = 1.
So, the values of x and y that maximize the product xy^3 are x = 60 and y = 1.
Checking the options:
a) x = 45, y = 15: xy^3 = 45 * 15^3 = 45 * 3375 = 151,875
b) x = 15, y = 45: xy^3 = 15 * 45^3 = 15 * 91,125 = 1,366,875
c) x = 10, y = 50: xy^3 = 10 * 50^3 = 10 * 125,000 = 1,250,000
d) x = 30, y = 30: xy^3 = 30 * 30^3 = 30 * 27,000 = 810,000
From the calculations, option B (x = 15, y = 45) gives the maximum value of xy^3, which is 1,366,875.
To find the values of x and y such that the product xy^3 is maximum, we need to maximize both x and y.
Given that x * y = 60, we can rewrite this equation as y = 60/x.
To maximize the product xy^3, we substitute the value of y from the above equation into the product:
P = x * (60/x)^3
Simplifying the equation, we have:
P = 60^3 / x^2
Maximizing x:
To maximize the product, we need to minimize the denominator, which means we need to maximize x.
To do this, we can take the derivative of P with respect to x and set it equal to zero to find the critical points:
dP/dx = -2 * 60^3 / x^3 = 0
Solving for x, we find x = 60.
Maximizing y:
Substituting the value of x = 60 into the equation y = 60/x, we find y = 1.
So, the values of x and y that maximize the product xy^3 are x = 60 and y = 1.
Checking the options:
a) x = 45, y = 15: xy^3 = 45 * 15^3 = 45 * 3375 = 151,875
b) x = 15, y = 45: xy^3 = 15 * 45^3 = 15 * 91,125 = 1,366,875
c) x = 10, y = 50: xy^3 = 10 * 50^3 = 10 * 125,000 = 1,250,000
d) x = 30, y = 30: xy^3 = 30 * 30^3 = 30 * 27,000 = 810,000
From the calculations, option B (x = 15, y = 45) gives the maximum value of xy^3, which is 1,366,875.
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Find two positive numbers x and y such that x + y = 60 andxy3 is maximuma)x = 45, y = 15b)x = 15, y = 45c)x = 10, y = 50d)x = 30, y = 30Correct answer is option 'B'. Can you explain this answer?
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