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Alik [6]
3 years ago
11

A parcel delivery service will deliver a package only if the length plus girth​ (distance around) does not exceed 84 inches. ​(A

) Find the dimensions of a rectangular box with square ends that satisfies the delivery​ service's restriction and has maximum volume. What is the maximum​ volume? ​(B) Find the dimensions​ (radius and​ height) of a cylindrical container that meets the delivery​ service's restriction and has maximum volume. What is the maximum​ volume?
Mathematics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

A. 14x14x28

B. The maximum volume is 5488 cuibic inches

Step-by-step explanation:

The problem states that the box has square ends, so you can express volume with:

v=x^{2} y

Using the restriction stated in the problem to get another equation you can substitute in the one above:

4x+y=84\\\\

Substituting <em>y</em> whit this equation gives:

v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}

Now find the limit of <em>x</em>:

\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14

Find the length:

y=84-4(14)=28

You can now calculate the maximum volume:

v=(14)^{2}(28)= 5488

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5. If 2x - y = 10 then for x= 8 what is the value of y?<br>a) 7<br>b) 6 <br>C) 5<br>d) 4​
sweet-ann [11.9K]
The answer is b, 6

if x=8 then you need to plug in 8 for your in the equation, so you’d have 2(8) - y = 10

you’d get: 16 - y = 10

you’d then subtract 16 from both sides to get -y alone

so now you have -y = -6

you can’t have a -y so you’ll have to divide -1 from both sides

you’ll end up with y=6, which is your answer
6 0
3 years ago
A circular bar is subjected to an axial pull of 100kN.if the maximum intensity of shear stress on any plane is not to exceed 60M
Oksana_A [137]
Use a Mohr circle to find the maximum shear stress relative to the axial stress.
Here we assume the axial stress is sigma, the transverse axial stress is zero.
So we have a Mohr circle with (0,0) and (0,sigma) as a diameter.
The centre of the circle is therefore (0,sigma/2), and the radius is sigma/2.
From the circle, we determine that the maximum stress is the maximum y-axis values, namely +/- sigma/2, at locations (sigma/2, sigma/2), and (sigma/2, -sigma/2).
Given that the maximum shear stress is 60 MPa, we have
sigma/2=60 MPa, or sigma=120 MPa.
(note: 1 MPa = 1N/mm^2)
Therefore
100 kN/(pi*d^2/4)=100,000 N/(pi*d^2/4)=120 MPa  where d is in mm.
Solve for d
d=sqrt(100,000*4/(120*pi))
=32.5735 mm
5 0
3 years ago
Evaluate. (-2 1/4)^2
balandron [24]

Answer:

5 \frac{1}{16}

Step-by-step explanation:

I did this before. Also sometimes once I look at the question I just find out the answer without steps.

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3 years ago
Find the slope of the line that passes through the given points:<br> (-18, 4), (-19, 18)
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Step-by-step explanation:

7 0
3 years ago
A total of 937 people attended the play admission was $2.00 for adults and $0.75 for students. The total ticket sales amounted t
mamaluj [8]

612 students attended and 325 adults attended the play

<h3><u>Solution:</u></h3>

Let "s" be the number of students attended the play

Let "a" be the number of adults attended the play

Cost of 1 adult ticket = $ 2

Cost of 1 student ticket = $ 0.75

<em><u>Given that total 937 people attended the play</u></em>

number of students + number of adults = 937

s + a = 937 ---------- eqn 1

<em><u>The total ticket sales amounted to $1,109</u></em>

number of adults x Cost of 1 adult ticket + number of students x Cost of 1 student ticket = 1109

a \times 2 + s \times 0.75 = 1109

2a + 0.75s = 1109 ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

s = 937 - a -------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

2a + 0.75(937 - a) = 1109

2a + 702.75 - 0.75a = 1109

1.25a = 1109 - 702.75

1.25a = 406.25

a = \frac{406.25}{1.25}

<h3>a = 325</h3>

From eqn 3,

s = 937 - 325

<h3>s = 612</h3>

Thus 612 students attended and 325 adults attended the play

3 0
3 years ago
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