Look At Me! Look At Me! Look At Me!
Shirley is designing a float for an upcoming Parade. She has a 20 meter long wire, that she wants to use to form the outline of a square and/or an equilateral triangle.
To attract the most attention from the parade attendees, she wants to maximize the sum of areas of these shapes. To 1 decimal place, what is the maximum area that she can cover?
Image credit: Wikipedia
The answer is 25.0.
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1 solution
This question is intentionally tricky, and I expect a lot of people to make mistakes on it (especially if they blindly differentiate).
Let the side length of the square be s and the side length of the triangle be t . We are given that 4 s + 3 t = 2 0 . Since 0 ≤ s and 0 ≤ t , we get the restriction that 0 ≤ s ≤ 5 .
We want to maximize the area, which is A = s 2 + 4 3 t 2 . Substituting in the previous equation, we get that
A = s 2 + 4 3 ( 3 2 0 − 4 s ) 2
Now, notice that this is a quadratic equation in s , with a positive leading coefficient, hence it open upwards. As such, the maximum will occur at one of the endpoints.
We can quickly check that the maximum occurs when s = 5 , which gives us an area of 5 2 = 2 5 . 0 .
Note: There is no need to expand out the quadratic equation. Those who differentiated would most likely have found the minimum value (aka the vertex of the parabola).