What's the area of these?

The top and bottom margins of a poster are 6 cm and the side margins are each 8 cm. If the area of printed material on the poste

Degger [83]

Answer:

The total height and width of the poster is approximately 28.93 and 38.57.

Step-by-step explanation:

Consider the provided information.

Let us consider the width and length of the printed part of the poster is x and y respectively.

For better understanding refer the figure 1:

The area of the printed part is 382 cm²

Area = xy = 382

y = 382/x

Now the total height of the poster is y + 6 + 6 = y + 12

Total width of the poster is x + 8 + 8 = x + 16

Thus the total area of the poster is:

A = (x + 16)(y + 12)

Now substitute the value of y in above equation.

$A = (x+16)(\frac{382}{x}+12)$

$A = 382+12x+\frac{6112}{x}+192$

$A = 12x+\frac{6112}{x}+574$

Now differentiate the above equation with respect to x.

$A' = 12-\frac{6112}{x^2}$

Now, substitute A'= 0 and solve for x.

$0 = 12-\frac{6112}{x^2}$

$12=\frac{6112}{x^2}$

$x^2=\frac{6112}{12}$

$x=22.57$ Ignore the negative value of x as width can't be a negative number.

Now find A''

$A'' = \frac{2(6112)}{x^3}$

Here A" is positive for x>0 and x = 22.57 is minimum.

Use the value of x to find all the respective dimensions of the poster.

Substitute the value of x in y = 382/x

y = 382/22.57 = 16.93

The total height of the poster is y + 6 + 6 = y + 12 = 16.93 + 12 = 28.93

Total width of the poster is x + 8 + 8 = x + 16 = 22.57 + 16 = 38.57

Hence, the total height and width of the poster is approximately 28.93 and 38.57.

6 0

3 years ago

Read 2 more answers

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

german

For this case we have the following function:
$f(x) = tan(\frac{x^2}{9})$
The average rate of change is given by:
$AVR = \frac{f(x2) - f(x1)}{x2- x1}$
Evaluating the function for the given interval we have:
For x = 1.25:
$f(1.25) = tan(\frac{1.25^2}{9})$
$f(1.25) = 0.18$
For x = 2:
$f(2) = tan(\frac{2^2}{9})$
$f(2) = 0.48$
Then, replacing values we have:
$AVR = \frac{0.48 - 0.18}{2 - 1.25}$
$AVR = 0.4$
Answer:
the average value of on the closed interval [1.25, 2] is:
$AVR = 0.4$

3 0

3 years ago

Solve for x. 3x/4 - X/12 =7/8 simplify your answer in fraction form

Usimov [2.4K]

Answer:

7/3

Step-by-step explanation:

or, 3x/4 - 7/8 = 7/8

or, 3x/4=7/8+7/8

or, 3x/4 = (7+7)/8

or, 3x/4 = 14/8

or, 3x = (14/8)*4