Can someone help me???

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago

NEED HELP !! There are 16 ounces in 1 pound . How many 12-ounce servings are in 36 pounds of dog food ? PLEASSSEEDE

dedylja [7]

There are (3) 12 ounce servings of dog food

7 0

3 years ago

Read 2 more answers

Given: ΔPSQ, PS = SQ Perimeter of ΔPSQ = 50 SQ – PQ = 1 Find: Area of ΔPSQ

Lana71 [14]

Answer:

120 units²

Step-by-step explanation:

Perimeter = PS + SQ + PQ

50 = SQ + SQ + (SQ -1)

51 = 3SQ

17 = SQ

17 -1 = 16 = PQ

The midpoint of the base is one leg of the right triangle whose other leg is the height of this isosceles triangle. That height is ...

h = √(17² -(16/2)²) = √225 = 15

Then the area is ...

A = (1/2)bh = (1/2)(16)(15) = 120 . . . . . square units

8 0

3 years ago

This triangle is a multiple of one of the triples listed here. What is the hypotenuse?

stiv31 [10]

Answer:

50 units

Step-by-step explanation:

you use Pythagorean-theorem which means A^2+B^2 =C^2

It does not matter the order of A and B because it is adding. But C is the hypotenuse.

If we plug in the # this is what it looks like...

14^2 +48^2=C^2

or

48^2+14^2=C^2

14*14= 196

48*48=2,304

then you add

196+2,304=C^2

2,500= C^2

and you can use a calculator for this part you do 2,500

with the 2,500 you use the sign as in the picture and it gives you...

50

which means 50*50 = 2,500

5 0

3 years ago

Read 2 more answers

-

alisha [4.7K]

Answer:

$d=3\sqrt{2}\ units$

Step-by-step explanation:

The complete question is

Line L contains points (3, 5) and (7, 9). Points P has coordinates (2, 10). Find the distance from P to L

step 1

Find the equation of the line L contains points (3,5) and (7,9)

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the given values

$m=\frac{9-5}{7-3}$

$m=\frac{4}{4}=1$

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=1\\point\ (3,5)$

substitute

$y-5=(1)(x-3)$

isolate the variable y

$y=x-3+5$

$y=x+2$ -----> equation A

step 2

Find the equation of the perpendicular line to the given line L that passes through the point P

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal

so

The slope of the given line L is $m=1$

The slope of the line perpendicular to the given line L is

$m=-1$

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=-1\\point\ (2,10)$

substitute

$y-10=-(x-2)$

isolate the variable y

$y=-x+2+10$

$y=-x+12$ ----> equation B

step 3

Find the intersection point equation A and equation B

$y=x+2$ -----> equation A

$y=-x+12$ ----> equation B

solve the system by graphing

The intersection point is (5.7)

see the attached figure

step 4

we know that

The distance from point P to the the line L is equal to the distance between the point P and point (5,7)

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

(2,10) and (5,7)

substitute

$d=\sqrt{(7-10)^{2}+(5-2)^{2}}$

$d=\sqrt{(-3)^{2}+(3)^{2}}$

$d=\sqrt{18}\ units$

simplify

$d=3\sqrt{2}\ units$

see the attached figure to better understand the problem

6 0

4 years ago