Precalc question

Question

An isosceles triangle has sides of length x, x and y. In addition, assume the triangle has perimeter 12.

(a) Find the rule for a function that computes the area of the triangle as a function of x. Describe the largest possible domain of this function.

(b) Assume that the maximum value of the function a(x) in (a) occurs when x = 4. Find the maximum value of z = a(x) and z = 2a(3x + 3) + 1.

Answer 1

Hi Flora,

The Cliff Notes version:

Step 1: We calculate the area using (1/2)base*height where height is found using pythagorean theorem:

A = (1/2)*y*sqrt(x^2 - (1/4)y^2)

Step 2: Use the given perimeter to express y in terms of x, then plug this expression into the equation for area to eliminate y:

A = (6 - x)*sqrt(12x - 36)

Step 3: Largest possible domain = all values of x for which the problem makes sense, i.e. x, x, y can form a legitimate triangle:

domain = (x_min, x_max) = (3,6)

Step 4: We are told that A(x) is max when x = 4 so plugging in x = 4 we get:

max value of z = A(4) = 4*sqrt(3)

Step 5: Finally, use the maximum area we just found to find max value of: 2A(3x + 3) + 1 = 2*4*sqrt(3) + 1 = 8*sqrt(3)+1

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The (really) long answer:

The formula for the area of any triangle is A=1/2(base)*(height). Since the triangle is isosceles, it seems natural to make the base y. Finding the height is a little tricker but can be computed using the pythagorean theorem. If we draw a vertical line down the middle of the triangle, we can break it up into two equal right triangles. The length of this vertical line is the height we're looking for.

Let's focus on one of these two equal triangles (it doesn't matter which one, they're equal). This right triangle has a hypotenuse, a vertical side, and a horizontal side. Let's call them, h, vs, and hs. From the pythagorean theorem, we know that the square of the hypotenuse (h) is equal to the sum of the squares of the two sides, so we can write this:

h^2 = vs^2 + hs^2

Since we are interested in the vertical side (vs), we can subtract hs^2 from both sides and rearrange to get the following:

vs^2 = h^2 - hs^2

Now, if draw this triangle (which I can't do here, unfortunately), you will notice that h = x and hs = 1/2(y). When we plug these values into the equation above, we get the following:

vs^2 = x^2 - (1/2y)^2 = x^2 - (1/4)y^2

Now, to solve for vs, we need to take the square root of both sides:

vs = sqrt(x^2 - (1/2)y^2)

which is the height of our isosceles triangle!

Now, we can use the area formula to calculate the area of the isosceles triangle.

A = 1/2(base)*(height)

We know that base = y and the height = vs = sqrt(x^2 - (1/4)y^2), so plugging in to our area equation, we get:

A = (1/2)*y*sqrt(x^2 - (1/4)y^2)

We're almost there. The perimeter is equal to 2x + y = 12. We can eliminate y from the area equation by solving for it in terms of x and plugging in again.

y = 12 - 2x

A = (1/2)*y*sqrt(x^2 - (1/4)y^2)

A = (1/2)*(12-2x)*sqrt(x^2 - (1/4)(12-2x)^2)

A = (6-x)*sqrt(x^2 - (1/4)*(144 - 48x+4x^2))

A = (6-x)*sqrt(x^2 - 36 +12x - x^2))

A = (6 - x)*sqrt(12x - 36)

Now to find the largest possible domain for this function.

The maximum value of x is 6 (which you can see from y = 12 - 2x)

The smallest value of x occurs when y = 2x which gives x = 3

Now we'll tackle part (b):

Since the area is largest at x = 4 (given in part b), we can plug 4 into our equation for the area to get the maximum value of z = A(x) = A(4).

z = A(4) = (6-4)*sqrt(12*4 - 36)

z = 2*sqrt(12) = 2*2*sqrt(3)

z = 4*sqrt(3)

Now for z = 2A(3x + 3)+1

z = 2*[(6 - 3x - 3)*sqrt(12*(3x + 3) - 36) + 1

z = (6 - 6x)*sqrt(36x + 36) +1

z = (6 - 6x)*6*sqrt(x+1)+1

z = 36*(1-x)*sqrt(x+1)+1

The maximum value of z in this case occurs when x = 0. Plugging in, we get:

The maximum value of z in this case occurs when A(x) = A(4) which means that x = 1/3. Plugging in, we get:

z = 2*4*sqrt(3)+1

z = 8*sqrt(3) + 1

Answer 2

2x+y=12

area=y/2*(c^2-a^2)^.5

In this case a=y and c=x within the ( ).

Answer 3

The perimeter equation is 2x + y = 12; therefore, y = 12 - 2x

The altitude h is such that h² + (y/2)² = x² so that h² = x² - (6-x)²

^{that is h = 2 sqrt(3x - 9)}

^{a(x) = (6-x) sqrt(3x - 9)}

^{The domain of a(x) is x>3 and a(x) does, in fact, have a maximum at x=4 and a(4) = 2 sqrt(3).}

^{I will leave the very last part for you to figure out.}