A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 120 in. What dimensions will give a box with a square end the largest possible volume?

20 in, 40 in

Step-by-step explanation:

Let L is the length and W is the width.

Total distance around, P = 2 L + 2 W = 120

L + W = 60 (1)

Volume of box, V = W² x L

V = W² x (60 - W)             from equation (1)

V = 60 W² - W³

dV/dW = 120 W - 3W²

It is equal to zero for maxima and minima

120 W - 3 W² = 0

W = 40 in

L = 60 - 40 = 20 in

Thus, the dimensions of the box is L = 20 in, W = 40 in

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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