How do I solve this ?

An open box is to be made from a 3 ft by 8 ft rectangular piece of sheet metal by cutting out squares of equal size from the fou

Burka [1]

Answer:

Maximum volume of the box will be 7.41 cubic feet.

Step-by-step explanation:

Open box has been made from a metal sheet measuring 3 ft and 8 ft.

Let four square pieces were removed from the four corners with one side measurement x ft.

Volume of the open box = Length × width × height

Length of the box = (3 - 2x) ft

Width of the box = (8 - 2x) ft

Height of the box = x ft

Volume of the box = (3 - 2x)(8 - 2x)x

V = (24 - 6x - 16x + 4x²)x

V = 24x - 22x² + 4x³

Now we take the derivative of V with respect to x and equate the derivative to zero,

$\frac{d}{dx}(V)=\frac{d}{dx}(24x - 22x^{2}+4x^{3})$

V' = 24 - 44x + 12x²

V' = 0

12x² - 44x + 24 = 0

3x² - 11x + 6 = 0

3x² - 9x - 2x + 6 = 0

3x(x - 3) - 2(x - 3) = 0

(3x - 2)(x - 3) = 0

(3x - 2) = 0

and (x - 3) = 0

Therefore, x = 3, $\frac{2}{3}$

For x = 0.67

Length of the box = (3 - 2x) = 3 - 1.34

= 1.66 ft

Width of the box = (8 - 2x) = 8 - 1.34

= 6.66 ft

Volume of the box = 0.67 × 1.66 × 6.66

V = 7.41 cubic feet.

Similarly, for x = 3,

Length of the box = (3 - 2\times 3) = -3

which is negative but the length of the box can not be negative.

Therefore, maximum volume of the box will be 7.41 cubic feet.

6 0

3 years ago

10. There are 355 students at Median Middle School. If there 8% of the students were absent on Tuesday, how many students were a

Setler79 [48]

To solve this question, we simply need to divide the total amount of students by 8% to find out how many students were absent and how many were present.

However, we can't simply multiply it by 8%, so we need to turn that into a decimal.

8% - 0.08

Multiply:

355 x 0.08

= 28.4

Round:

28 kids were absent

Subtract:

355 - 28

= 327

This means that 28 kids were absent, and 327 kids were present.

6 0

2 years ago

Please Help!!

algol13

When you see questions of this nature, test the individual inequalities and look out for their intersection.

For
$y < \frac{2}{3} x$
Choose a point in the lower or upper half plane created by the line
$y = \frac{2}{3} x$
The above line is the one which goes through the origin.

Now testing (1,0) yields,

$0 < \frac{2}{3} (1)$
That is,

$0 < \frac{2}{3}$
This statement is true. So we shade the lower half of
$y = \frac{2}{3} x$

For
$y \geqslant - x + 2$
We test for the origin because, it is not passing through the origin.

$0 \geqslant - (0) + 2$
This yields
$0 \geqslant 2$
This statement is false so we shade the upper half.

The intersection is the region shaded in B. The top right graph

8 0

4 years ago

My brothers and sisters i need help asap please

Otrada [13]

Answer:

Step-by-step explanation:

I think it is D but I am not sure

4 0

3 years ago

Como resolver logaritmos avanzados

Firlakuza [10]

Answer:

could u put this in english

5 0

3 years ago