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# the top and bottom margins of a poster are each cm and the side margins are each cm. if the area of printed material on the poster is fixed at , find the dimensions of the poster with the smallest area.

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### James

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get the top and bottom margins of a poster are each cm and the side margins are each cm. if the area of printed material on the poster is fixed at , find the dimensions of the poster with the smallest area. from EN Bilgi.

## The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm.

CALCULUS Simon G. asked • 07/22/20

## The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm.

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm2, find the dimensions of the poster with the smallest area.

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Since we are being asked to find the smallest area, we know that we are trying to find a minimum of the area function.

Let's let "W" be the overall width of poster and "H" be the overall height of the poster.

A = H•W

But we also have information that will help us get the number of variables down to one. We know that the area of the printed material is 384. The width of the printed material is W - 4(2) or W - 8. The height of the printed material is H - 6(2) = H - 12. So the area of the printed material is (W - 8)(H - 12) and that equals 384

(W - 8)(H - 12) = 384

WH - 12W - 8H + 96 = 384

W(H - 12) - 8H = 288

W = (288 + 8H)/(H - 12)

Plugging that into the area equation (A = H•W), we get:

A(H) = H(288 + 8H)/(H - 12)

A(H) = (288H + 8H2)/(H - 12)

To find the minimum, take the derivative and set it equal to zero.

To take the derivative, we must use the quotient rule

If f(x) = u/v then f '(x) = (u'v - uv')/v2

In this case u = 288H + 8H2 so u' = 288 + 16H

and v = H - 12 so v' = 1 and v2 = (H - 12)2 so

A'(H) = [(288 + 16H)(H - 12) - (288H + 8H2)(1)]/(H - 12)2

A'(H) = (288H - 3456 + 16H2 - 192H - 288H - 8H2)/(H - 12)2

A'(H) = (8H2 - 192H - 3456)/(H - 12)2

A'(H) = 8(H2 - 24H - 432)/(H - 12)2

Setting it equal to zero:

8(H2 - 24H - 432)/(H - 12)2 = 0

(H2 - 24H - 432)/(H - 12)2 = 0 This can only be zero if the numerator is zero so:

H2 - 24H - 432 = 0

(H - 36)(H + 12) = 0

H = 36 or H = -12

We can throw out the negative answer since the poster can't be a negative height so H = 36. Since H = 36, and we know that W = (288 + 8H)/(H - 12), we can solve for W:

W = (288 + 8(36))/(36 - 12)

W = 24 So the poster has W = 24 cm H = 36 cm Upvote 0 Downvote Comments 2 More Jeff K.

Here's how to do it without solving any quadratics! Using the same notation: (W-8)(H-12) = 384 Which means: W = 384/(H-12) + 8 . . . . . . . eqn (1) So, total area = A = WH = H [384/((H-12) + 8] By the chain rule, dA/dH = 1.[(384/(H-12) +8] + H[-384/(H-12)^2] For min/max area, dA/dH = 0 => 384/(H-12) + 8 - 384H/(H-12)^2 = 0 Common denominator is (H-12)^2: 384(H-12) + 8(H-12)^2 - 384H = 0 We can multiply both sides by the denominator, (H-12)^2, since it can't be zero. Can you see why from the original diagram? Therefore: 384H - 384 x 12 + 8(H - 12)^2 - 384H = 0 The first and last terms on the left cancel to zero, so we are left with: 8(H - 12)^2 = 384 x 12 (H - 12)^2 = 384 x 12 / 8 (H - 12)^2 = 48 x 12 (H - 12) = sqrt(48 x 12) = 24 . . . . . [taking sq roots on both sides Therefore, H = 12 + 24 = 36 Substitute into eqn (1) above to get W = 16 + 8 = 24 Solution set: (W, H) = (24, 36)

Report 07/22/20 Jeff K.

Sorry, the system removed all the line breaks. -:-( Hope you can still read it.

Report 07/22/20

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## The top and bottom margins of a poster are 4 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 384 square centimeters, how do you find the dimensions of the poster with the smallest area?

Draft Let a be the width of the poster and b the height. Let A be the area of the poster to be minimized. (In this explanation I will omit all the "cm"). A=384+2(color(red)(a*4))+2(color(blue)(b*6))-4(color(orange)(6*4))=384+8a+12b-96=288+8a+12b As the sum of the illustrated parts: A=a*b So let's get a in function of b: 288+8a+12b=ab 288+12b=ab-8a 288+12b=a(b-8) a=(288+12b)/(b-8) Now, A(b) will be the function in one single variable (b) that we will minimize: A(b)=a*b=((288+12b)/(b-8))*b=(288b+12b^2)/(b-8). I have to find the first derivative of the function to minimize it: A'(b)=12(b^2-16b-192)/(b-8)^2 The minimum points satisfy the condition A'(b)=0, so: b^2-16b-192=0 b=-8 or b=24 But for obvious reason (b is the height of a poster), b must be positive, so only b=24 is a correct solution. Now we have to find a: a=(288+12b)/(b-8)=(288+12*24)/(24-8)=36 So, the solution to the problem is (a,b)=(24,36).

## The top and bottom margins of a poster are 4 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 384 square centimeters, how do you find the dimensions of the poster with the smallest area?

Calculus Applications of Derivatives Solving Optimization Problems

Luca F. Jun 10, 2015 Draft

### Explanation:

Let a be the width of the poster and b the height.

Let A be the area of the poster to be minimized.

(In this explanation I will omit all the "cm").

A = 384 + 2 ( a ⋅ 4 ) + 2 ( b ⋅ 6 ) − 4 ( 6 ⋅ 4 ) = 384 + 8 a + 12 b − 96 = 288 + 8 a + 12 b

As the sum of the illustrated parts:

A = a ⋅ b

So let's get a in function of b:

288 + 8 a + 12 b = a b 288 + 12 b = a b − 8 a 288 + 12 b = a ( b − 8 ) a = 288 + 12 b b − 8 Now, A ( b )

will be the function in one single variable (b) that we will minimize:

A ( b ) = a ⋅ b = ( 288 + 12 b b − 8 ) ⋅ b = 288 b + 12 b 2 b − 8 .

I have to find the first derivative of the function to minimize it:

A ' ( b ) = 12 b 2 − 16 b − 192 ( b − 8 ) 2

The minimum points satisfy the condition

A ' ( b ) = 0 , so: b 2 − 16 b − 192 = 0 b = − 8 or b = 24

But for obvious reason (b is the height of a poster), b must be positive, so only b=24 is a correct solution.

Now we have to find a:

a = 288 + 12 b b − 8 = 288 + 12 ⋅ 24 24 − 8 = 36

So, the solution to the problem is

( a , b ) = ( 24 , 36 ) . Answer link

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## The top and bottom margins of a poster are each 6 cm and the

Find step-by-step Calculus solutions and your answer to the following textbook question: The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm , find the dimensions of the poster with the smallest area..

### Question

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm , find the dimensions of the poster with the smallest area.

### Explanation

Verified Let x,y

x,y be the width and height respectively of the printed area, which has fixed area

xy = 384 xy=384 cm ^2 2

\implies \qquad y = \dfrac{ 384}{x }

⟹y= x 384 ​

The total height of the poster including 6 cm margins at the top and bottom is

y + 12 y+12.

The total width of the poster including 4 cm margins at sides is

x + 8 x+8.

The area of the total poster is

A = (x+8)(y+12) A=(x+8)(y+12)

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A rectangular poster is to contain 648 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used?

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A rectangular poster is to contain 648 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used?

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James 9 month ago

Guys, does anyone know the answer?