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

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SUMMARY

The maximum of the function f(x) = 12x - 32 + 24√(9 - 3x) for x ≤ 3 is determined by analyzing the function's critical points and endpoints. The function is defined for x values up to 3, where the square root remains valid. By evaluating f(x) at the endpoint x = 3 and finding the derivative, the maximum value is confirmed to occur at x = 3, yielding f(3) = 12. Thus, the maximum value of f(x) is 12.

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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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