ms sue
posted by Riana .
triangle base is (4x)cm and its height is (2x+3)cm.what will be the maximium area?

ms sue  I don't know 
Ms. Sue
Sorry.

ms sue 
Reiny
Area = (1/2)base x heigh
= (1/2)(4x)(2x+3)
= (1/2)(12 + 5x  2x^2)
d(Area)/dx = (1/2)(5  4x)
= 0 for a max area
4x = 5
x = 5/4
using x = 5/4 = 1/25
Maximum Area = (1/2)(45/4)(5/2+3) = 7.5625
or appr 7.5625
testing: our answer t= 5/4 = 1.25
let x = 1.24 , area = (1/2)(41.24)(2.48+3) = 7.5624 , a bit smaller
let x = 1.26, area = (1/2)(41.26)(2.52+3) = 7.5624 , a bit smaller again
all looks good at max area = 7.5625 
ms sue 
Riana
can you explain how you got 1/2 x 54x ?

ms sue 
Reiny
Since this is a typical Calculus question, I assume you knew Calculus.
I took the derivative. 
ms sue 
Reiny
You could start with my area equation
area = (1/2)(12 + 5x  2x^2)
= (1/2)(2)(x^2  5/2x 6)
= (x^2  (5/2)x + 25/16  25/16  6) I competed the square
= ( (x  5/4)^2  25/16  6)
= (x  5/4)^2 + 25/16 + 6
= (x5/4)^2 + 121/16
area = (x5/4)^2 + 121/16
max area = 121/16 or 7.5625
when x = 5/4
just as in my first solution.
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