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4 years ago

Consider the point.

Mathematics

1 answer:

aalyn [17]4 years ago

8 0

Answer:

L = 8.366

Step-by-step explanation:

Given

A(0,0,0)

H(3,5,6)

we can apply the equation

L = √((xH-xA)²+(yH-yA)²+(zH-zA)²)

where L is the length of the diagonal of the box

then

L = √((3-0)²+(5-0)²+(6-0)²)

⇒ L = √(9+25+36) = √70

⇒ L = 8.366

We can see the box in the pic shown.

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3 years ago

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.

8 0

4 years ago