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3 years ago

6

Julie had three wheels from bikes and things that she stacked against the shed. Each wheel fitted so neatly together that Julie

took this photo. The radius of the largest sheet is 16 cm and the radius of the middle-sized wheel is 9cm. What is the radius of Julie’s smallest wheel?

1 answer:

Lady bird [3.3K]3 years ago

3 0

Answer:

the answer will be 2 cm

Step-by-step explanation:

As the difference between 1st and 2nd wheel's radius is 7 cm

16cm-9cm= 7cm

therefore the difference between the middle-wheel and smallest wheel also should be 7 cm, lets assume smallest wheel radius = x

9cm-x= 7cm

x=9-7

x= 2 cm

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A farmer has 300 ft of fence to enclose 2 adjacent rectangular pens bordering his barn. Find the maximum area he can close.

Margarita [4]

Answer:

Width of 37.5 feet and length of 50 feet will maximize the area.

Step-by-step explanation:

Let w represent width and l represent length of the pen.

We have been given that a farmer has 300 ft of fence to enclose 2 adjacent rectangular pens bordering his barn. We are asked to find the dimensions that will maximize the area.

We can see from the attachment that the perimeter of the pens would be $4w+3l$ .

We can set our given information in an equation $4w+3l=300$ .

The area of the two pens would be $A=2l\cdot w$ .

From perimeter equation, we will get:

$w=\frac{300-3l}{4}$

Substituting this value in area equation, we will get:

$A=2l\cdot (\frac{300-3l}{4})$

Since we need to maximize area, so we need to find derivative of area function as:

$A=\frac{600l-6l^2}{4}$

Bring out the constant:

$A=\frac{1}{4}*\frac{d}{dl}(600l-6l^2)$

$A=\frac{1}{4}*(600-12l)$

$A=150-3l$

Now, we will set our derivative equal to 0 as:

$150-3l=0$

$150=3l$

$\frac{150}{3}=\frac{3l}{3}$

$50=l$

Now, we will substitute $l=50$ in equation $w=\frac{300-3l}{4}$ to solve for width as:

$w=\frac{300-3(50)}{4}$

$w=\frac{300-150}{4}$

$w=\frac{150}{4}$

$w=37.5$

Therefore, width of 37.5 feet and length of 50 feet will maximize the area.

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