Lol... I don't know if this is a joke, but it's basically just restating the precious problem, but switching numbers. So the answer would be 2,592.
Answer:
1
Step-by-step explanation:
Amplitude is the greatest distance from the mean line/mid line
Answer:
The answer is D - 3.786x10^7
Answer:
Step-by-step explanation:
We have the equations
4x + 3y = 18 where x = the side of the square and y = the side of the triangle
For the areas:
A = x^2 + √3y/2* y/2
A = x^2 + √3y^2/4
From the first equation x = (18 - 3y)/4
So substituting in the area equation:
A = [ (18 - 3y)/4]^2 + √3y^2/4
A = (18 - 3y)^2 / 16 + √3y^2/4
Now for maximum / minimum area the derivative = 0 so we have
A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0
-3/8 (18 - 3y) + √3 y /2 = 0
-27/4 + 9y/8 + √3y /2 = 0
-54 + 9y + 4√3y = 0
y = 54 / 15.93
= 3.39 metres
So x = (18-3(3.39) / 4 = 1.96.
This is a minimum value for x.
So the total length of wire the square for minimum total area is 4 * 1.96
= 7.84 m
There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.
