MHB How to Find the Maximum of the Function f(x)=12x-32+24√(9-3x), for x≤3?

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To find the maximum of the function f(x) = 12x - 32 + 24√(9 - 3x) for x ≤ 3, first identify the domain, which is limited by the square root, leading to x values between negative infinity and 3. The function is continuous and differentiable within this domain, allowing for the application of calculus to find critical points. By taking the derivative and setting it to zero, critical points can be determined, and evaluating the function at these points and the endpoint x = 3 will reveal the maximum value. The maximum occurs at either a critical point or the endpoint, depending on the calculations. Ultimately, the maximum value of f(x) can be established through these evaluations.
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$f(x)=12x-32+24\sqrt {9-3x}, x\leq 3$
find $max f(x)$
 
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Albert said:
$f(x)=12x-32+24\sqrt {9-3x}, x\leq 3$
find $max f(x)$

let 9-3x = y^2 so 3x = 9-y^2 so $f(x) = 36-4y^2 - 32 + 24y = 4 - (4y^2 - 24 y) = 4 - 4(y- 3)^2 + 36 = 40 - 4(y-3)^2$
lowest when y = 3 ( range of y is >0 feasible and x = 0 so i we get 40 as maximum $f(x)$
 

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