Math / Algebra

Question19) ( section 10.6) A rectangular page is designed to contain 64 square inches of print. The margins at the top and bottom of the page are each 1 inch deep. The margins on each side are $11 / 2$ inches wide. Find an algebraic expression for the area of the entire page. Using a calculator find the dimensions that would create the minimum area. (approx. 2 decimal places)

Studdy Solution

STEP 1

What is this asking? We need to find a formula for the total area of a page with some print on it, then figure out the best page dimensions to use the least amount of paper while still holding all that print! Watch out! Don't mix up the area of the printed part with the total area of the page!
Also, remember those margins are on both sides, top and bottom!

STEP 2

1. Set up variables and express the printed area.
2. Express the total page area.
3. Optimize the total area.
4. Calculate the dimensions.

STEP 3

Let $x$ be the width and $y$ be the height of the printed area.
We know the printed area is 64 square inches, so we can write this relationship as: $x \cdot y = 64$

STEP 4

We can solve for $y$ in terms of $x$ by dividing both sides by $x$ : $y = \frac{64}{x}$ This will be useful later!

STEP 5

The printed area has a width of $x$ .
We have two side margins, each $1\frac{1}{2} = \frac{3}{2}$ inches wide.
So the total page width is $x + 2 \cdot \frac{3}{2} = x + 3$ inches.

STEP 6

The printed area has a height of $y$ .
We have top and bottom margins of 1 inch each.
So the total page height is $y + 1 + 1 = y + 2$ inches.

STEP 7

The total area, which we'll call $A$ , is just width times height: $A = (x+3)(y+2)$ Now, let's substitute $y = \frac{64}{x}$ from the first step: $A = (x+3)(\frac{64}{x} + 2)$ Distribute to get: $A = 64 + 2x + \frac{192}{x} + 6$ $A = 2x + \frac{192}{x} + 70$ Awesome! Now we have the total area in terms of just one variable, $x$ !

STEP 8

To minimize the area, we need to find where the derivative of $A$ with respect to $x$ is zero.
Let's find the derivative: $\frac{dA}{dx} = 2 - \frac{192}{x^2}$

STEP 9

Now, let's set it to zero and solve for $x$ : $2 - \frac{192}{x^2} = 0$ $2 = \frac{192}{x^2}$ $2x^2 = 192$ $x^2 = 96$ $x = \sqrt{96}$ Since $x$ represents a width, it must be positive, so we take the positive square root.

STEP 10

We found $x = \sqrt{96} \approx 9.80$ inches.

STEP 11

Now, let's find $y$ : [ y = \frac{64}{x} = \frac{64}{\sqrt{96}} = \frac{64}{4\sqrt{6}} = \frac{16}{\sqrt{6}} = \frac{16\sqrt{6}}{6} = \frac{8\sqrt{6}}{3} \approx 6.53 \) inches.

STEP 12

[Finally, the total page width is $x + 3 = \sqrt{96} + 3 \approx 12.80$ inches, and the total page height is $y + 2 = \frac{8\sqrt{6}}{3} + 2 \approx 8.53$ inches.

STEP 13

[The dimensions that minimize the total area of the page are approximately 12.80 inches wide by 8.53 inches tall.
The algebraic expression for the area of the entire page is $A = 2x + \frac{192}{x} + 70$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up variables and express the printed area.
2. Express the total page area.
3. Optimize the total area.
4. Calculate the dimensions.

STEP 3

Let $x$ be the width and $y$ be the height of the printed area.
We know the printed area is 64 square inches, so we can write this relationship as: $x \cdot y = 64$

STEP 4

We can solve for $y$ in terms of $x$ by dividing both sides by $x$ : $y = \frac{64}{x}$ This will be useful later!

STEP 5

The printed area has a width of $x$ .
We have two side margins, each $1\frac{1}{2} = \frac{3}{2}$ inches wide.
So the total page width is $x + 2 \cdot \frac{3}{2} = x + 3$ inches.

STEP 6

The printed area has a height of $y$ .
We have top and bottom margins of 1 inch each.
So the total page height is $y + 1 + 1 = y + 2$ inches.

STEP 7

STEP 8

To minimize the area, we need to find where the derivative of $A$ with respect to $x$ is zero.
Let's find the derivative: $\frac{dA}{dx} = 2 - \frac{192}{x^2}$

STEP 9

Now, let's set it to zero and solve for $x$ : $2 - \frac{192}{x^2} = 0$ $2 = \frac{192}{x^2}$ $2x^2 = 192$ $x^2 = 96$ $x = \sqrt{96}$ Since $x$ represents a width, it must be positive, so we take the positive square root.

STEP 10

We found $x = \sqrt{96} \approx 9.80$ inches.

STEP 11

Now, let's find $y$ : [ y = \frac{64}{x} = \frac{64}{\sqrt{96}} = \frac{64}{4\sqrt{6}} = \frac{16}{\sqrt{6}} = \frac{16\sqrt{6}}{6} = \frac{8\sqrt{6}}{3} \approx 6.53 \) inches.

STEP 12

[Finally, the total page width is $x + 3 = \sqrt{96} + 3 \approx 12.80$ inches, and the total page height is $y + 2 = \frac{8\sqrt{6}}{3} + 2 \approx 8.53$ inches.

STEP 13

[The dimensions that minimize the total area of the page are approximately 12.80 inches wide by 8.53 inches tall.
The algebraic expression for the area of the entire page is $A = 2x + \frac{192}{x} + 70$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up variables and express the printed area.2. Express the total page area.3. Optimize the total area.4. Calculate the dimensions.

STEP 3

Let xxx be the width and yyy be the height of the **printed area**.We know the printed area is **64 square inches**, so we can write this relationship as: x⋅y=64 x \cdot y = 64 x⋅y=64

STEP 4

We can solve for yyy in terms of xxx by dividing both sides by xxx: y=64x y = \frac{64}{x} y=x64​ This will be useful later!

STEP 5

The printed area has a width of xxx.We have two side margins, each 112=321\frac{1}{2} = \frac{3}{2}121​=23​ inches wide.So the total page width is x+2⋅32=x+3 x + 2 \cdot \frac{3}{2} = x + 3 x+2⋅23​=x+3 inches.

STEP 6

The printed area has a height of yyy.We have top and bottom margins of 1 inch each.So the total page height is y+1+1=y+2 y + 1 + 1 = y + 2 y+1+1=y+2 inches.

STEP 7

STEP 8

To minimize the area, we need to find where the derivative of AAA with respect to xxx is zero.Let's **find the derivative**: dAdx=2−192x2 \frac{dA}{dx} = 2 - \frac{192}{x^2} dxdA​=2−x2192​

STEP 9

Now, let's **set it to zero** and solve for xxx: 2−192x2=0 2 - \frac{192}{x^2} = 0 2−x2192​=0 2=192x2 2 = \frac{192}{x^2} 2=x2192​2x2=192 2x^2 = 192 2x2=192x2=96 x^2 = 96 x2=96x=96 x = \sqrt{96} x=96​Since xxx represents a width, it must be positive, so we take the positive square root.

STEP 10

We found x=96≈9.80 x = \sqrt{96} \approx 9.80 x=96​≈9.80 inches.

STEP 11

Now, let's find yyy: [ y = \frac{64}{x} = \frac{64}{\sqrt{96}} = \frac{64}{4\sqrt{6}} = \frac{16}{\sqrt{6}} = \frac{16\sqrt{6}}{6} = \frac{8\sqrt{6}}{3} \approx 6.53 \) inches.

STEP 12

[Finally, the **total page width** is x+3=96+3≈12.80 x + 3 = \sqrt{96} + 3 \approx 12.80 x+3=96​+3≈12.80 inches, and the **total page height** is y+2=863+2≈8.53 y + 2 = \frac{8\sqrt{6}}{3} + 2 \approx 8.53 y+2=386​​+2≈8.53 inches.

STEP 13

[The dimensions that minimize the total area of the page are approximately **12.80 inches** wide by **8.53 inches** tall.The algebraic expression for the area of the entire page is A=2x+192x+70A = 2x + \frac{192}{x} + 70A=2x+x192​+70.

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1. Set up variables and express the printed area.
2. Express the total page area.
3. Optimize the total area.
4. Calculate the dimensions.

Let $x$ be the width and $y$ be the height of the printed area.
We know the printed area is 64 square inches, so we can write this relationship as: $x \cdot y = 64$

We can solve for $y$ in terms of $x$ by dividing both sides by $x$ : $y = \frac{64}{x}$ This will be useful later!

The printed area has a width of $x$ .
We have two side margins, each $1\frac{1}{2} = \frac{3}{2}$ inches wide.
So the total page width is $x + 2 \cdot \frac{3}{2} = x + 3$ inches.

The printed area has a height of $y$ .
We have top and bottom margins of 1 inch each.
So the total page height is $y + 1 + 1 = y + 2$ inches.

To minimize the area, we need to find where the derivative of $A$ with respect to $x$ is zero.
Let's find the derivative: $\frac{dA}{dx} = 2 - \frac{192}{x^2}$

Now, let's set it to zero and solve for $x$ : $2 - \frac{192}{x^2} = 0$ $2 = \frac{192}{x^2}$ $2x^2 = 192$ $x^2 = 96$ $x = \sqrt{96}$ Since $x$ represents a width, it must be positive, so we take the positive square root.

We found $x = \sqrt{96} \approx 9.80$ inches.

Now, let's find $y$ : [ y = \frac{64}{x} = \frac{64}{\sqrt{96}} = \frac{64}{4\sqrt{6}} = \frac{16}{\sqrt{6}} = \frac{16\sqrt{6}}{6} = \frac{8\sqrt{6}}{3} \approx 6.53 \) inches.

[Finally, the total page width is $x + 3 = \sqrt{96} + 3 \approx 12.80$ inches, and the total page height is $y + 2 = \frac{8\sqrt{6}}{3} + 2 \approx 8.53$ inches.

[The dimensions that minimize the total area of the page are approximately 12.80 inches wide by 8.53 inches tall.
The algebraic expression for the area of the entire page is $A = 2x + \frac{192}{x} + 70$ .