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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Maximum
Albert1
Messages
1,221
Reaction score
0
$f(x)=12x-32+24\sqrt {9-3x}, x\leq 3$
find $max f(x)$
 
Mathematics news on Phys.org
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 $f(x)$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
9
Views
3K
Back
Top