sum of height and base is 58... maximum area?

Question

The sum of the height h and the base b of a triangle is 58. What height and base will produce a triangle of maximum area?

h=?
b=?

Answer 1

A = (1/2)(b)(h)
h + b = 58

Subtract b from both sides of second equation:
h = 58 - b
And plug into first equation:
A = (1/2)(b)(58-b) = 29b - (b^2)/2

Put into ax^2+bx+c form:
-(1/2)b^2 + 29b + 0
Find the vertex (maximum point):
b = "-b/2a" = -29 / 2(-1/2) = 29

So the base is 29
h + b = 58
h + 29 = 58
h = 29

So the base and height are both 29.

Answer 2

Let's create a function for the height, given the base.

Let b = base
Let h = 58 - b:

A = ½ b*h
A = ½(b)(58 - b)
A = -½b² + 29b
Multiply both sides by -2 to get rid of the -½:
-2A = b² - 58b

Now complete the square by taking the coefficient on the b term (-58), divide it in half (-29) and square it (841). Add it and subtract it:
-2A = b² - 58b + 841 - 841

The first part is a perfect square:
-2A = (b² - 58b + 841) - 841

-2A = (b - 29)² - 841

Now divide by -½:
A = -½(b - 29)² + 420½

Notice that the first part is subtracting a square. A square will never be smaller than 0. So you want this to be 0 so you take away the least from the area.

(b-29)² = 0
(b-29) = 0
b = 29

So when the base is 29 and the height is 29, the triangle has a maximum area of 420½ sq. units.

Answer 3

The triangle with the greatest area is always a right triangle, with equal height and base. (Seeing as the rectangle with the greatest area is a square, also with equal height and base)
Thus, height and base are both 29.

Answer 4

A = bh/2
b + h = 58
A= (h-58)(h/2)
A= ((h^2)/2)-29h
to get the maximum area: use derivative
A' = (2h)(1/2) - 29 = 0
h-29=0
h = 29
b = 58 - 29 = 29