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If x and y are positive numbers satisfying 3x 4y=34 then maximum value of x³y²?
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If x and y are positive numbers satisfying 3x 4y=34 then maximum value...
Finding the Maximum Value of x³y²


Given the equation 3x + 4y = 34, we need to find the maximum value of x³y². To do this, we need to use some optimization techniques.


Step 1: Express y in terms of x


We can rearrange the equation to get y in terms of x as follows:

4y = 34 - 3x

y = (34 - 3x)/4


Step 2: Substitute y into the expression for x³y²


Now we can substitute the expression for y into the expression for x³y²:

x³y² = x³[(34 - 3x)/4]² = x³(1156 - 204x + 9x²)/16


Step 3: Find the derivative of x³y² with respect to x


To find the maximum value of x³y², we need to find the value of x that makes the derivative of x³y² with respect to x equal to zero. We can find the derivative using the product rule:

(x³(1156 - 204x + 9x²)/16)' = 3x²(1156 - 204x + 9x²)/16 + x³(18x - 204)/16

Simplifying this expression, we get:

(x³(27x² - 408x + 1156))/16


Step 4: Solve for x


Now we can set the derivative equal to zero and solve for x:

x³(27x² - 408x + 1156) = 0

x(x² - 12x + 43)(27x² - 408x + 1156) = 0

The solutions to this equation are x = 2.23, x = 5.51, and x = 2.16. However, only x = 2.16 and x = 5.51 are positive.


Step 5: Find the maximum value of x³y²


Now we can substitute these values of x back into the expression for x³y² and find the maximum value:

When x = 2.16, x³y² = 16.44

When x = 5.51, x³y² = 437.76

Therefore, the maximum value of x³y² is 437.76.
Community Answer
If x and y are positive numbers satisfying 3x 4y=34 then maximum value...
The only way you can do it is....
3×6 + 4×4 =34 so x=6 & y=4 so the maximum value of X^3y^2 is (6×6×6)×(4×4) = 3456
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If x and y are positive numbers satisfying 3x 4y=34 then maximum value of x³y²?
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