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.