Posted by **Riana** on Monday, January 28, 2013 at 7:31pm.

triangle base is (4-x)cm and its height is (2x+3)cm.what will be the maximium area?

- ms sue - I don't know -
**Ms. Sue**, Monday, January 28, 2013 at 7:44pm
Sorry.

- ms sue -
**Reiny**, Monday, January 28, 2013 at 8:55pm
Area = (1/2)base x heigh

= (1/2)(4-x)(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)(4-5/4)(5/2+3) = 7.5625

or appr 7.5625

testing: our answer t= 5/4 = 1.25

let x = 1.24 , area = (1/2)(4-1.24)(2.48+3) = 7.5624 , a bit smaller

let x = 1.26, area = (1/2)(4-1.26)(2.52+3) = 7.5624 , a bit smaller again

all looks good at max area = 7.5625

- ms sue -
**Riana**, Monday, January 28, 2013 at 9:21pm
can you explain how you got 1/2 x 5-4x ?

- ms sue -
**Reiny**, Monday, January 28, 2013 at 9:27pm
Since this is a typical Calculus question, I assume you knew Calculus.

I took the derivative.

- ms sue -
**Reiny**, Monday, January 28, 2013 at 9:40pm
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

= -(x-5/4)^2 + 121/16

area = -(x-5/4)^2 + 121/16

max area = 121/16 or 7.5625

when x = 5/4

just as in my first solution.

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