A poster is to have a total area of 180 cm^2 . There is a margin round the edges of 6cm at the top and 4 cm at the sides and bottom where nothing is printed.

Question

A poster is to have a total area of 180 cm^2 . There is a margin round the edges of 6cm at the top and 4 cm at the sides and bottom where nothing is printed. 

What width should the poster be in order to have the largest printed area? 

The poster should have width__________  cm

Answer 1

First, let's be careful with what the problem asks for, which is the width of the poster, not the width of the printed area. So we can express the total area as

w·h = 180,

which allows us to express h in terms of w:

h = 180/w

Now let's express the printed area:

a_p = (w - 8)(h - 10), or

a_p = (w - 8)(180/w - 10)

Multiply this out and clean-up:

a_p = 260 - 10w - 1440/w

Now the key thing to keep in mind is that a function's maxima and minima occur when its slope is zero. The slope of the printed area is found from the derivative:

a^'_p = 0 - 10 + 1440/w²

So when the slope is zero:

-10 + 1440/w² = 0

1440/w² = 10

144 = w², or just

w = 12

Answer 2

The poster is two rectangles:

1) The inner "printed" rectangle, whose area will be A_P

2) The larger total rectangle which includes the blank margin, whose area will be A_T

If we consider that the length of the printed area is L and the width is W, we can start with an equation for the printed area of the poster, which is just the area of a rectangle:

A_P = LW

Our goal is to maximize A_P, which means we need to take a derivative and set it equal to zero to find the critical points. However, in order to do that, we need to get A_P in terms of a single variable. We're going to need a second equation so we can substitute something in. That's where the area of the larger rectangle comes in. Since the margins for the poster are 4 cm on the sides, we can say that the length of the larger rectangle is L+8 (i.e. W plus two 4 cm margins). Similarly, the width is W+10, since the top margin is slightly bigger. So the area of the larger rectangle is:

A_T = (L+8)(W+10)

We also know that A_T is supposed to be 180, so the equation becomes:

180 = (L+8)(W+10)

Foil this out, subtract 8w and 80 from the right side, factor out L, and divide by (W+10), and we get:

https://bit.ly/3aV8QYH

We can now substitute this into our equation for A_P:

https://bit.ly/2wWN7Rs

Now it's just a matter of differentiating the equation and setting it equal to zero:

https://bit.ly/2TO3uJ0

We can do this derivative using the quotient rule:

https://bit.ly/3d2rPlQ

https://bit.ly/2TOzpJw

Phew. Once that's done, we set this new expression equal to zero so we can find the critical points:

https://bit.ly/2WeBQXt

This is a rational equation. To find the zeros of this thing, we really just need to consider the zeros for the numerator (since any fraction is equal to zero only when its numerator is equal to zero). This isn't too hard, since the numerator is just a quadratic:

-8W^2 - 160W + 1000 = 0

Dividing by -8 gives:

W^2 + 20W - 125 = 0

(W + 25)(W - 5) = 0

And therefore, the two zeros are W = -25 and W = 5. Since the width can't be negative, we throw that solution in the garbage, and we're left with W = 5. We can sub that value into our equation for L to find the length:

https://bit.ly/33gM3nG

So we get that the width of the printed area is W=5 and the length of the printed area is L=4. For the whole poster, we need to add 8 to the length and 10 to the width, giving the total length of 12 and total width of 15.

***Note: I typed most of this out using W for the vertical dimension, and it just occurred to me that "width" is probably referring to the horizontal dimension, which I called L. So the answer is probably 12 cm. That's also a really small printed area, which makes me a little uneasy.

Answer 3

This is a constrained optimization problem (i.e. we want to maximize the area of the printed poster subject to the contraint of the total poster area). As such, we want to approach it using the Lagrange multiplier method.

First of all, we want to write our constraint equation:

(1) (h + 10)(w + 8) = 180

This equation comes from adding the height of the poster plus the 6 cm on top and 4 cm on the bottom and multiplying by the width plus the 4 cm on either side. Then, we want to write our maximization equation:

(2) f(w, h) = wh

This is just the formula for the area of the poster itself. From this, we can define our Lagrangian:

(3) L(w, h, λ) = wh + λ(wh + 10w + 8h - 100)

Equation (3) comes from adding our function to lambda multiplied by the expanded out constraint equation where we have set the constraint equal to 0 (by moving over the 180 to the lefthand side). Next, we have three first order conditions:

(4) 0 = dL/dw = h + λh + 10λ

(5) 0 = dL/dh = w + λw + 8λ

(6) 0 = dL/dλ = wh + 10w + 8h -100

Let us start with equations (4) and (5). Moving over the h and w to the lefthand side and dividing the equations followed by cross multiplication yields:

(7) -h/-w = (λh + 10λ) / (λw + 8λ)

(8) h/w = (h + 10)/(w + 8)

(9) hw + 8h = hw + 10w

(10) h = 5/4 w

We can then plug this value of h into our constaint (6) which yields:

(11) 5/4w² +10w + 10w - 100 = 0

(12) 5/4w² + 20w - 100 = 0

We can then apply the quadratic formula to find that

(13) w = (-20 +/- sqrt(400 - 4*5/4*-100)) / (2 * 5/4)

(14) w = (-20 +/- sqrt(900) )/(2.5)

(15) w = (-20 +/- 30)/(2.5)

(16) w = 10 / 2.5

(17) w = 4

Where in step 15, we selected the positive option in the quadratic formula as our width must be positive. Thus, we get a width of 4 to maximize our area. We could plug this back into our constraint (1) to solve for the height (h + 10)(12) = 180, so h = 5. This means in turn that the maximum area is 5 * 4 = 20. We could go and check the second order conditions to verify that this is in fact a maximum, but we'll just assume that here. I hope that helps!

Answer 4

Okay--this is a maximization problem given a constraint.

We have our function to be maximized f(x,y) [[here f(x,y) = A(x,y) = xy to be maximized)

We have our constraint function (shown below) we call g(x,y)

g(x,y) is the dimensions of the poster = 180 = (y+10)(x+8)

vertical y + 6" + 10" margins, horiz + two 4" margins

We set "Solution= S(x,y) = f(x,y) - λg(x,y)

We take dS/dx = dS/dy = 0

(((SEE BELOW AT "SOLUTION")))

********************************************************************************************************************

We want to maximize the area of say X*Y=A within the border of the poster which has 6" margin atop, and 4" margins L, R, and below--if I interpreted correct.

Let X be our imaged horizontal and y, our vertical.

Hence

f(x,y) = xy

g(x,y) = (x+8)(y+10) = 180 = xy + 10x + 8y + 80

(Do you get this?) IF SO--SKIP TO BELOW THE ILLUSTRATION

*********************************************************************************************

The same result can be found by adding the squares and rectangles of the image

(crappy job here but does trick---

----------------------------

| 6^ |

| ____x___ |

| 4 | | 4 |

| | (y)| |

| |________| |

| 4 |

---------------------------

First the CORNER SQUARES OF THE MARGIN

View upper left square of 4x6, upper right of 4x6 (all in cm squared)

lower left and lower right both of 4x4----this gives 24+24+16+16 or 48+32 or

**(80)

NOW THE SIDES AND TOP AND BOTTOM RECTANGLES: We have atop "x" by 6;

at left and right we have "y" by 4, and at bottom we have "x" by 4--this gives

**4x+6x+4y+4y= 10X+8Y

Now we add the central quadrilateral: area of

X*Y

Adding these we get XY+ 10X+8Y+80=180 which is equivalent to (x+8)(y+10)=180 (see?)

****************************************************************************************************************

SOLUTION

So we have "S(x,y)" = f(x,y) - λ g(x,y)

then dS/dx = 0 = dS/dy

or xy - λ(xy+10x+8y-100)

or xy - λxy - 10xλ - 8yλ -100λ

TAKE DERIVATIVE w/rspct to x:

y - λy - 10λ = 0 OR λ = y / (y+10)

TAKE DERIVATIVE w/rspct to y:

x - λx - 8λ = 0 OR λ = x / (x+8)

NOW SET λs equal: x/(x+8) = y / (y+10)

x(y+10) = y(x+8)

xy + 10x = xy + 8y

10x = 8y

1.25x=y

Substitute this into (x+8)(y+10)=180

or xy + 10x + 8y - 100=0 to get

1.25x² + 10x+ 8*(1.25x) -100 = 0

Solve using (a,b,c) = (1.25, 20,-100) to get X=4 Y =5

The area is 20; the dimensions are as noted.