A rain gutter is made from sheets of aluminum that are 12 inches wide by turning up the edges to form right angles. Determine the depth of the gutter that will maximize its cross-sectional area and allow the greatest amount of water to flow. What is the maximum cross-sectional area?

Depth = 3 inches

Maximum cross-sectional area = 18 inches²

Step-by-step explanation:

Let 'D' be the depth of the gutter and 'W' be the width of the gutter, the cross-sectional area as function of depth, A(D), is:

The depth for which the derivate of the area function is zero is the depth that yields the maximum cross-sectional area:

The cross-sectional area for D = 3 is:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.