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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 5

Lesson 11: Solving optimization problems

# Optimization: area of triangle & square (Part 1)

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.B (LO)
,
FUN‑4.B.1 (EK)
,
FUN‑4.C (LO)
,
FUN‑4.C.1 (EK)
Sal constructs an equilateral triangle & a square whose bases are 100m together, such that their area is the smallest possible. Created by Sal Khan.

## Want to join the conversation?

• Would it not have been much easier (and faster) to recognize a 30-60-90 triangle at and use simple geometry to realize the sides are at ratios of 1, 2, sqrt(3) -> as S/2, S, (S/2)*Sqrt(3)?
• Absolutely. But you can learn to trust a new method by first ensuring that it is consistent with other methods in cases where you can apply the other method. Then when you are outside of a familiar context, you can apply the more general method.
• h can be found way faster by using s * sin(60)

I don't understand why this long method is shown, what is the advantage?
• The method shown can be adapted to a variety of triangles and quadrilaterals, so it is good to know a general method. You can only use the sine if you know the angle.

So, while use of the sine might have made this particular problem a little simpler, it wouldn't help in a more general case. Though, I would concur that it would be good to know both methods, so that you can choose whatever happens to be easier for a particular problem.
• Instead of using his "aside" with the triangle beginning at , I did something similar but using x/3 instead of 'S'. In other words, I said:
Using the Pythagorean Theorem: ((x/3)/2)^2+h^2 = (x/3)^2
h^2 = (x/3)^2-(x/6)^2
h = x/3-x/6
h=x/6

Obviously this differs from Sal's answer but I can't figure out why?
• Your mistake was when you square rooted,
√(a²+b²) is NOT √a² + √b² and is not a+b
Here is the correct way to do it:
h² = (⅓x)²- (⅙x)²
h² = ¹⁄₉ x² - ¹⁄₃₆ x²
h² = ¹⁄₁₂ x²
h = x / √12
h = x / (2√3)
h = ⅙ x√3
• would you use the same formula for an isoscles triangle
• No. Area is still 1/2 s * h but it has a base length a, and two sides length b. So a + 2b = x/3.
• How would you do this same problem but with a circle instead of a triangle?
• The area of a circle is pi * r^2. The circumference, which is x in this case, will be 2 * pi * r. If we calculate for r we get that r = x / (2 * pi). If we substitute r in the formula for the area we get Area = pi * ( x / (2 * pi) )^2 = x^2 / (4 * pi)
• Near the beginning of the video, Sal decides arbitrarily to call "x" the length of the left-hand section of wire (the one that the triangle will be formed from). What if we chose to call "x" the length of the side that the square was made out of? It seems to me that how you choose to define "x" and which shape you make out of which section ("x" vs "100-x") affects the character of the function used to find the combined area. If the problem didn't stipulate which side the square was made from and which side the triangle was made from, wouldn't you have to work through the entire problem both ways in order to be sure that you'd effectively minimized it?
• You should get the same result either way. If you're minimising the same quantity, then the values you arbitrate on the way shouldn't make any difference.
• i am so lost @ for the combined area of the triangle and square we added the two area's together. why is he multiplying by (x/3)^2 when he already found the area to be ( squareroot of 3/ 4 ) ?
• The area is not √3/4
It's √3/4 s²
And he determined earlier that
s = x/3

In this problem there are some unknowns. You aren't trying to calculate the area as a number because you may not have enough information to do that initially.
So Sal is creating a function for finding the area.

It may help to watch the next video in which he calculates the actual area.
• At when Sal says," We are going to minimize the combined area of the triangle and the square" Does he mean he wants to find the smallest possible area of the square and triangle combined?
• Yep, he wants the total area to be as small as possible.
• why don't you use the equation 1/2 ab sin(c) to find the area of the triangle? is it not the same answer?
• Sal could have used 1/2 ab sin(C), and would have still gotten sqrt(3)/4 * s^2. I guess he just felt like using the other formula, as 1/2 ab sin(C) is a relatively less known formula.
• I paused the video and tried it my way. Not sure if it's correct though.

I set the sides of the triangle to x and the sides of the square to y. From there, I set the combined area as, f(x) = √3/4 x² + y².

I related x and y with the equation 100 = 3x + 4y. From there, I solved for y, and substituted it in for y in f(x). Then I found the critical points and used the 2nd derivative test. Is my method correct?
(1 vote)
• When you do optimization problem, you want to have 1 variable. So instead of calling x and y. You can call one piece of the wire x, then the remaining piece must be 100-x.
Now when you make an equilateral triangle from x, then each side is not x but x/3. So the area is (√3s²)/4 and s=x/3, similarly, the side of the square is going to be the second piece divide into 4, or (100-x)/4, and the area is s² or [(100-x)/4)²

Then take the derivative and find the critical number, then use either first or second derivative test of your choice. However, this is a close interval, [0,100], So you want to include the lower bound, 0, and the upper bound, 100 (in other words, endpoints) as your critical numbers.