Math / Algebra

QuestionDiana has 440 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area? (a) $A(W)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle given a fixed perimeter, then figure out the biggest area we can make and what rectangle width gets us there. Watch out! Don't mix up perimeter and area!
Also, remember the width affects the length, they're linked!

STEP 2

1. Set up the perimeter equation
2. Express length in terms of width
3. Define the area function
4. Optimize the area formula
5. Calculate the maximum area

STEP 3

Alright, let's start with what we know: Diana has 440 yards of fencing.
That's the perimeter of our rectangle!

STEP 4

Remember, the perimeter of a rectangle is two times the length plus two times the width.
Let's write that down: $2L + 2W = 440$ Where $L$ is the length and $W$ is the width.

STEP 5

We want the area as a function of only the width, $W$ .
So, let's get $L$ out of the perimeter equation!

STEP 6

First, let's divide the entire perimeter equation by 2: $L + W = 220$

STEP 7

Next, let's subtract $W$ from both sides to isolate $L$ : $L = 220 - W$ Boom! Now we have $L$ in terms of $W$ !

STEP 8

The area of a rectangle is length times width, right?
So, let's write that down: $A = L \cdot W$

STEP 9

But we just found a cool way to write $L$ using only $W$ !
Let's substitute that in: $A(W) = (220 - W) \cdot W$

STEP 10

Let's expand this to make it look even nicer: $A(W) = 220W - W^2$ There it is!
Our area function in terms of the width!

STEP 11

Now, how do we find the largest area?
This is where our vertex formula comes in handy!
Remember, the x-coordinate of the vertex of a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

STEP 12

In our area function, $a = -1$ and $b = 220$ , and $W$ acts like our $x$ .
So, the width that gives us the maximum area is: $W = -\frac{220}{2 \cdot (-1)}$

STEP 13

Calculate that out: $W = 110$ So, a width of 110 yards will give us the biggest rectangle!

STEP 14

Let's plug our magic width, $W = 110$ , back into our area function: $A(110) = 220 \cdot 110 - (110)^2$

STEP 15

Calculate that: $A(110) = 24200 - 12100$

STEP 16

Final calculation: $A(110) = 12100$ The maximum area is 12,100 square yards!

STEP 17

(a) $A(W) = 220W - W^2$ (b) $W = 110$ yards (c) Maximum area = 12,100 square yards

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QuestionDiana has 440 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area? (a) $A(W)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle given a fixed perimeter, then figure out the biggest area we can make and what rectangle width gets us there. Watch out! Don't mix up perimeter and area!
Also, remember the width affects the length, they're linked!

STEP 2

1. Set up the perimeter equation
2. Express length in terms of width
3. Define the area function
4. Optimize the area formula
5. Calculate the maximum area

STEP 3

Alright, let's start with what we know: Diana has 440 yards of fencing.
That's the perimeter of our rectangle!

STEP 4

Remember, the perimeter of a rectangle is two times the length plus two times the width.
Let's write that down: $2L + 2W = 440$ Where $L$ is the length and $W$ is the width.

STEP 5

We want the area as a function of only the width, $W$ .
So, let's get $L$ out of the perimeter equation!

STEP 6

First, let's divide the entire perimeter equation by 2: $L + W = 220$

STEP 7

Next, let's subtract $W$ from both sides to isolate $L$ : $L = 220 - W$ Boom! Now we have $L$ in terms of $W$ !

STEP 8

The area of a rectangle is length times width, right?
So, let's write that down: $A = L \cdot W$

STEP 9

But we just found a cool way to write $L$ using only $W$ !
Let's substitute that in: $A(W) = (220 - W) \cdot W$

STEP 10

Let's expand this to make it look even nicer: $A(W) = 220W - W^2$ There it is!
Our area function in terms of the width!

STEP 11

Now, how do we find the largest area?
This is where our vertex formula comes in handy!
Remember, the x-coordinate of the vertex of a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

STEP 12

In our area function, $a = -1$ and $b = 220$ , and $W$ acts like our $x$ .
So, the width that gives us the maximum area is: $W = -\frac{220}{2 \cdot (-1)}$

STEP 13

Calculate that out: $W = 110$ So, a width of 110 yards will give us the biggest rectangle!

STEP 14

Let's plug our magic width, $W = 110$ , back into our area function: $A(110) = 220 \cdot 110 - (110)^2$

STEP 15

Calculate that: $A(110) = 24200 - 12100$

STEP 16

Final calculation: $A(110) = 12100$ The maximum area is 12,100 square yards!

STEP 17

(a) $A(W) = 220W - W^2$ (b) $W = 110$ yards (c) Maximum area = 12,100 square yards

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QuestionDiana has 440 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area? (a) A(W)=A(W)=A(W)= □\square□

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the area of a rectangle given a fixed perimeter, then figure out the biggest area we can make and what rectangle width gets us there. Watch out! Don't mix up perimeter and area!Also, remember the width affects the length, they're linked!

STEP 2

1. Set up the perimeter equation2. Express length in terms of width3. Define the area function4. Optimize the area formula5. Calculate the maximum area

STEP 3

Alright, let's **start** with what we know: Diana has **440 yards** of fencing.That's the **perimeter** of our rectangle!

STEP 4

Remember, the perimeter of a rectangle is *two times the length* plus *two times the width*.Let's write that down: 2L+2W=440 2L + 2W = 440 2L+2W=440 Where LLL is the length and WWW is the width.

STEP 5

We want the area as a function of *only* the width, WWW.So, let's get LLL out of the perimeter equation!

STEP 6

**First**, let's divide the entire perimeter equation by **2**: L+W=220 L + W = 220 L+W=220

STEP 7

**Next**, let's subtract WWW from both sides to isolate LLL: L=220−W L = 220 - W L=220−W Boom! Now we have LLL in terms of WWW!

STEP 8

The **area** of a rectangle is length times width, right?So, let's write that down: A=L⋅W A = L \cdot W A=L⋅W

STEP 9

But we just found a cool way to write LLL using only WWW!Let's **substitute** that in: A(W)=(220−W)⋅W A(W) = (220 - W) \cdot W A(W)=(220−W)⋅W

STEP 10

Let's **expand** this to make it look even nicer: A(W)=220W−W2 A(W) = 220W - W^2 A(W)=220W−W2 There it is!Our area function in terms of the width!

STEP 11

Now, how do we find the *largest* area?This is where our **vertex formula** comes in handy!Remember, the x-coordinate of the vertex of a parabola y=ax2+bx+c y = ax^2 + bx + c y=ax2+bx+c is given by x=−b2a x = -\frac{b}{2a} x=−2ab​.

STEP 12

In our area function, a=−1 a = -1 a=−1 and b=220 b = 220 b=220, and WWW acts like our xxx.So, the width that gives us the maximum area is: W=−2202⋅(−1) W = -\frac{220}{2 \cdot (-1)} W=−2⋅(−1)220​

STEP 13

**Calculate** that out: W=110 W = 110 W=110 So, a width of **110 yards** will give us the biggest rectangle!

STEP 14

Let's **plug** our magic width, W=110W = 110W=110, back into our area function: A(110)=220⋅110−(110)2 A(110) = 220 \cdot 110 - (110)^2 A(110)=220⋅110−(110)2

STEP 15

**Calculate** that: A(110)=24200−12100 A(110) = 24200 - 12100 A(110)=24200−12100

STEP 16

**Final calculation**: A(110)=12100 A(110) = 12100 A(110)=12100 The **maximum area** is **12,100 square yards**!

STEP 17

(a) A(W)=220W−W2 A(W) = 220W - W^2 A(W)=220W−W2 (b) W=110 W = 110 W=110 yards (c) Maximum area = 12,100 square yards

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QuestionDiana has 440 yards of fencing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width W of the rectangle. (b) For what value of W is the area largest? (c) What is the maximum area? (a) $A(W)=$ $\square$

What is this asking? We need to find a formula for the area of a rectangle given a fixed perimeter, then figure out the biggest area we can make and what rectangle width gets us there. Watch out! Don't mix up perimeter and area!
Also, remember the width affects the length, they're linked!

1. Set up the perimeter equation
2. Express length in terms of width
3. Define the area function
4. Optimize the area formula
5. Calculate the maximum area

Alright, let's start with what we know: Diana has 440 yards of fencing.
That's the perimeter of our rectangle!

Remember, the perimeter of a rectangle is two times the length plus two times the width.
Let's write that down: $2L + 2W = 440$ Where $L$ is the length and $W$ is the width.

We want the area as a function of only the width, $W$ .
So, let's get $L$ out of the perimeter equation!

First, let's divide the entire perimeter equation by 2: $L + W = 220$

Next, let's subtract $W$ from both sides to isolate $L$ : $L = 220 - W$ Boom! Now we have $L$ in terms of $W$ !

The area of a rectangle is length times width, right?
So, let's write that down: $A = L \cdot W$

But we just found a cool way to write $L$ using only $W$ !
Let's substitute that in: $A(W) = (220 - W) \cdot W$

Let's expand this to make it look even nicer: $A(W) = 220W - W^2$ There it is!
Our area function in terms of the width!

Now, how do we find the largest area?
This is where our vertex formula comes in handy!
Remember, the x-coordinate of the vertex of a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

In our area function, $a = -1$ and $b = 220$ , and $W$ acts like our $x$ .
So, the width that gives us the maximum area is: $W = -\frac{220}{2 \cdot (-1)}$

Calculate that out: $W = 110$ So, a width of 110 yards will give us the biggest rectangle!

Let's plug our magic width, $W = 110$ , back into our area function: $A(110) = 220 \cdot 110 - (110)^2$

Calculate that: $A(110) = 24200 - 12100$

Final calculation: $A(110) = 12100$ The maximum area is 12,100 square yards!

(a) $A(W) = 220W - W^2$ (b) $W = 110$ yards (c) Maximum area = 12,100 square yards