A poster is to have an area of 210in^2 with 1 inch margins at the bottom and sides and a 2 inch margin at the top. Find the exact dimensions that will give the largest printed area.

Let's denote the width of the poster as "x" inches and the height of the printed area as "h" inches.

We know that the total area of the poster is equal to the area of the printed area plus the sum of the margins:
Area of the poster = Area of the printed area + Sum of margins

In this case, the area of the printed area is equal to the area of the poster minus the sum of the margins:
Area of the printed area = Area of the poster - Sum of margins

The area of the printed area is given as 210 square inches.

We can calculate the sum of the margins as follows:
Sum of margins = (2 x x) + (1 x h) + (1 x h) = 2x + 2h

Now we can substitute the given values into the equation for the area of the printed area:
210 = (x x h) - (2x + 2h)

Simplifying the equation gives us:
xh = 210 + 2x + 2h

Now we want to find the exact dimensions that will give the largest printed area, which means we want to maximize the area of the printed area in terms of x and h.

To do this, we can express the equation in terms of one variable and then use calculus to find the maximum. Let's solve the equation for h:

xh = 210 + 2x + 2h
Subtract 2h from both sides:
xh - 2h = 210 + 2x
Factor out h:
h(x - 2) = 210 + 2x
Divide both sides by (x - 2):
h = (210 + 2x) / (x - 2)

Now we have an expression for h in terms of x. To find the maximum area, we can take the derivative of the area of the printed area with respect to x and set it equal to zero:
d/dx (xh) = d/dx (x((210 + 2x) / (x - 2))) = 0

Using the quotient rule, we can find the expression for the derivative:
((x - 2)(2) - (210 + 2x)(1)) / (x - 2)^2 = 0

Simplifying the expression gives us:
-210 + 4 - x + 2x = 0
x - 206 = 0
x = 206

Now we can substitute this value of x into the expression for h to find the corresponding value of h:
h = (210 + 2(206)) / (206 - 2) = 622 / 204 = 3.04

Therefore, the exact dimensions that will give the largest printed area are approximately 206 inches for the width and 3.04 inches for the height.

To find the dimensions that will give the largest printed area for the poster, we will need to maximize the area while taking into account the given margins.

Let's start by understanding the dimensions of the poster. We know that it has 1-inch margins at the bottom and sides, and a 2-inch margin at the top. Let's denote the width of the printed area as "w" and the height of the printed area as "h".

Based on this information, we can write the following equations:

1. The width of the poster: w = w + (2 * 1) inches
(The width of the printed area plus the sum of the margins on both sides)

2. The height of the poster: h = h + (2 * 1) + 2 inches
(The height of the printed area plus the sum of the margins on the bottom, sides, and top)

Now, let's calculate the total area of the poster using the given information:

Total area = (width of the poster) * (height of the poster)
= w * h

Since we want to maximize the area, we need to find the dimensions that will give us the maximum value for w * h.

To do this, we can express one of the variables in terms of the other and then find the maximum value of the total area.

From equation 1, we can rewrite w as follows:
w = w + 2 inches
w - w = 2 inches
0 = 2 inches
This equation is not solvable because 0 cannot equal 2 inches.

This means that width w cannot be expressed in terms of w, and we need to approach the problem in a different way.

Since the total area is given by the product of the width and height, we can express either the width or height in terms of a single variable and then maximize the value of the total area.

Let's express the height h in terms of the width w:

From equation 2, we can rearrange the equation as follows:
h = h + 4 inches
h - h = 4 inches
0 = 4 inches
This equation is not solvable because 0 cannot equal 4 inches.

Again, we encounter a similar issue where we cannot express the height h in terms of the width w.

In this case, we should consider another approach: expressing one of the variables in terms of the other. Let's express the width w in terms of the height h:

From equation 1, we can rearrange the equation as follows:
w = w + 2 inches
w - w = 2 inches
0 = 2 inches
This equation is not solvable because 0 cannot equal 2 inches.

We face the same issue again: we can't express the width w in terms of the height h.

At this point, it seems like there might be an error in the given information, so it's worth double-checking or seeking clarification.

However, if we assume that the margins are meant to be equal on all sides, we can solve the problem with adjusted dimensions.

Let's denote the width of the printed area without the margins as "w'" and the height of the printed area without the margins as "h'".

Based on the adjusted dimensions, the width w' would be w' = w - 2 inches, and the height h' would be h' = h - 4 inches.

Now we can calculate the total area:

Total area = (width of the printed area) * (height of the printed area)
= (w - 2 inches) * (h - 4 inches)

We want to maximize the total area, so let's find the exact dimensions that will do so.

210 = (h+3) (L +2) = h L + 3 L + 2 h + 6

or
h L + 3 L + 2 h = 204
L(h+3) = 204-2h
L = (204-2h)/(h+3)

A = L h
A = h(204-2h)/(h+3)

for max dA/dh = 0
= h[(h+3)(-2) -204+2h]/(h+3)^2 +(204-2h)/(h+3)

=h[-2h-6-204+2h]/(h+3)^2 +(204-2h)/(h+3)

=-210 h/(h+3)^2 + (204-h)(h+3)/(h+3)^2
= zero for max
so numerator = 0
0 = -210 h + 204 h -h^2 +612-3h

0 = -h^2 -9 h + 612

h^2 + 9 h - 612 = 0

h = [ -9 +/- sqrt (81+2448) ]/2

h = [ -9 +/- 50.3 ]/2

h = 20.64
L = (204-2h)/(h+3)
= 6.89
check arithmetic