find L and W. My question is down

Question

An advertisement consists of a rectangular printed region plus 4-cm margins on the sides and 5-cm margins at top and bottom. If the area of the printed region is to be 186 cm2, find the dimensions of the printed region that minimize the total area.

Answer 1

Let width be sideways on the paper and length be up and down. Let T_W and T_L represent total width and total length and let W and L represent width and length of the printed area. Then T_W = W + 8 and T_L = L + 10.

We are trying to minimize total area so we need a function of total area that we can differentiate.

Total Area (A) is the product of total width and total length.

A = (T_W)(T_L)

A = (W + 8)(L + 10)

but we are told printed are is 186 cm² so W•L=186 therefore W = 186/L making our total area equation say:

A = (186/L + 8)(L + 10)

A = 186 + 1860/L + 8L + 80

A(L) = 8L + 1860L^-1 + 266

Take the derivative:

A’(L) = 8 - 1860L^-2

Set equal to zero and solve:

0 = 8 - 1860L^-2

1860/L² = 8

1860 = 8L²

232.5 =L²

L = 15.25 cm

Plugging this into W = 186/L gives W = 12.20 cm

Answer 2

xy = 186

x = 186/y

x and y = the horizontal and vertical lengths of the printed matter. x+8 and y+10 are the horizontal and vertical lengths of the total poster area

the full poster is (x+8)(y+10)

replace x with 186/y

Total Area = (186/y +8)(y+10) = 186 + 8y +1860/y +80

A = 266 +8y +1860/y = 266+8y + 1860y^-1

take the derivative and set = to zero

A' =8 - 1860/y^2 = 0

8y^2 -1860 = 0

y^2 - 232.5 = 0

y = sqr232.5 = about 15.245 cm

x = 186/y = 186/15.25 = about 12.200 cm

dimensions of the printed matter are 12.2 by 15.246 cm

which minimize the total postal area

12.2 times 15.246 = about 186.001 cm^2

total poster area dimensions are 20.2 by 25.246 cm

total area = 20.2 times 25.246 = about 509.67 cm^2 = minimum area for the poster

for 12 by 15.5, printed area = 186 cm^2

poster dimensions 20 by 25.5 with total area = 510 cm^2

for 12.4 by 15, printed area =186 cm^2

poster dimensions 20.4 by 25 with total area = 510 cm^2

if you rounded off the 15.246 cm to 15.25 cm

then the total area of the poster would be 20.2 times 25.25 = 510.05 which is greater than dimesnions 20.4 by 25 or 20 by 25.5, each with total area 510 cm^2

it's a very thin, fine difference, but 15.25 by 12.2 cm for the printed matter does not give the minimum total area

Answer 3

Printed area:

A=LW=186

W=186/L

Total Area:

T = (W+8)(L+10) = (186/L+8)(L+10)

T(L) = 186+1860/L+8L+80

T'(L) =-1860/L²+8

Set T'(L) to zero and solve for L

1860/L²=8

L=√(1860/8) ≈ 15.25 cm

W=186/15.25 ≈12.2 cm