# ms sue

posted by on .

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 - ,

Sorry.

• ms sue - ,

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 - ,

can you explain how you got 1/2 x 5-4x ?

• ms sue - ,

Since this is a typical Calculus question, I assume you knew Calculus.
I took the derivative.

• ms sue - ,

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.