The graph shows the function f(x).

Mathematics

1 answer:

strojnjashka [21]3 years ago

4 0

-2 that's the answers

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A piece of wire 8 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral trian

padilas [110]

Answer:

a. 8

b. 7

Step-by-step explanation:

Let $x$ be the length of wire for the square and $y = 8 -x$ (eq. 1) be the length of the wire for the equilateral triangle.

We know the $x$ is the total length of the wire used, hence the length of a side of the square will be $\frac{x}{4}$ . The area of the square will be: $(\frac{x}{4}) ^ 2$

Similarly for the equilateral triangle, perimeter is $y$ . Hence the area will be $\frac {\sqrt {3}\ y^2}{4}$ .

The total area of both shapes will be: $A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ y^2}{4}$

We will substitute the value of y from eq. 1:

$A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ (8-x)^2}{4}$

We find the derivative of the above function to find maximum and minimum: $f'(x) = 0$ ⇒ $A'(x) = 0$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2}$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2} = 0\\\\\frac{x}{8} - \frac{8\sqrt{3} -\sqrt{3} x}{2} =0\\\\x - 4(8\sqrt{3} -\sqrt{3}x)=0\\\\x-32\sqrt3 - 4\sqrt3 x = 0\\\\x - 4\sqrt3 = 32 \sqrt3 \\\\(1 - 4\sqrt3)x = 32 \sqrt3 \\\\x = \frac{32 \sqrt3 }{(1 - 4\sqrt3)}\\\\x =6.99 \approx 7.00$