Answer:
(a) A(x) = (12 -2x)√(3x -9)
Explanation:
You want a formula for the area (a(x)) of an isosceles triangle with congruent sides of length x and perimeter 12. You also want its maximum value, and a graph and the maximum value of z = 2a(3x+3)+1.
(a) Heron's formula
Heron's formula for the area of a triangle with side lengths a, b, c is ...
A = √(s(s -a)(s -b)(s -c)) . . . . . . where s = (a+b+c)/2, the semiperimeter
Given a perimeter of 12, and two sides x, this becomes ...
A(x) = √(6(6 -x)(6 -x)(6 -(12-2x))) = √(6(6 -x)²(2x -6))
A(x) = 2(6 -x)√(3(x -3)) . . . . . . factor out perfect squares
A(x) = (12 -2x)√(3x -9) . . . . . . domain: 3 ≤ x ≤ 6
The domain is such that the radical and the area are non-negative.
(b) Maximum
The derivative of the area function is ...
The derivative is zero at the maximum, so ...
-9x +36 = 0 ⇒ x = 4 . . . . . . . an equilateral triangle
The maximum area is ...
A(4) = (12 -2·4)√(3·4 -9) = 4√3
The maximum area is 4√3.
Since the maximum of A(x) is 4√3 at x=4, the maximum of 2A(3x+3) +1 will be ...
2(4√3) +1 = 8√3 +1 . . . . . . . where 3x+3 = 4 ⇒ x = 1/3
The maximum of 2A(3x+3)+1 is 8√3 +1 at x = 1/3.
(c) Scaled and translated
The function ...
z = 2A(3x+3) +1 = 2A(3(x+1)) +1
Compare this to ...
z = p·A(q(x -r)) +s
which is vertically expanded by p, horizontally compressed by q, then translated right r and up s.
We see that the parent function A(x) is vertically expanded by a factor of 2, compressed horizontally by a factor of 3, then translated left 1 and up 1.
Applying these transformations to the original A(x) domain and range, we find the domain is ...
1/3[3, 6] -1 = [0, 1] . . . . . domain of z
And the range is ...
2[0, 4√3] +1 = [1, 8√3 +1] . . . . . range of z