You are planning to make an open rectangular box from a 33-in.-by-65-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume?

Volume = 6783.27inch³

H = 6.951inch

L = 51.098inch

W = 19.098inch

Step-by-step explanation:

From the given information:

Let h = side length and height of the box

L = length of the box

W = width of the box

V = volume of the box

We have that:

L = 65-2h

W = 33-2h

V = L×W×h

Therefore we have

V = (65-2h)(33-2h)×h

V = (2145-130h-66h+4h²)×h

V = 2145h-196h²+4h³

By differentiating V w.r.t h, we have

V' = 2145-392h+12h²

V = 12h²-392h+2145

Using Almighty formula we have

h = 392+/-√392²-4(12)(2145)/2(12)

h = 6.951 or 25.716

Thus, we find L,W and V.

We use the least value of h in order not to get a negative value of volume, length and width

L = 65-2h = 65-2(6.951) = 51.098

W = 33-2h = 33-2(6.951) = 19.098

V = LWH = 51.098×19.098×6.951 = 6783.2696inches

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

answer: '

the distance between points l and m

step-by-step explanation: