Calculus- Minimum Problem?

Question

each page of a book will contain 30in^2 of print, and each page must have 2in margins at the top and bottom of the page. and 1 in margin on the left and right.
What is the minimum possible area of such a page?

Answer 1

let the length of the page of print=x
let the breadth of the page of print=y
area of page in print=30
so, x*y=30
y=30/x
the total length of the page will be=length of page in print+margin at top+margin at bottom=x+4

similarly,total breadth of page=y+2

area of whole page=length * breadth
area =(x+4)*(y+2)
substitute value of y
area = (x+4)*((30/x)+2)
differentiate both sides with respect to x.
since area has to be minimum its differentiation will be equal to 0.
so, 0=d/dx((x+4)*((30/x)+2).
0 = d/dx(30+2x+120/x+8)
0=2-120/x^2
x^2 = 60
x = 7.75inches
substitute the x in first eq to find y
y=30/x
y = 30/7.75 =3.87inches
then area = (x+4)*(y+2)
=(7.75+4)*(3.87+2)
=(11.75)*(5.87)
=68.98 sq inches (ans)

Answer 2

Let
x = width of print
y = height of print
A = area of page

We have:

xy = 30
y = 30/x

A = (x + 2)(y + 4)
A = (x + 2)(30/x + 4)
A = 30 + 4x + 60/x + 8
A = 38 + 4x + 60/x

Take the derivative and set equal to zero to find critical points.

dA/dx = 4 - 60/x² = 0
4 = 60/x²
4x² = 60
x² = 15
x = √15
The negative solution is rejected since x must be positive.

Calculate the area by plugging in for x.

A = 38 + 4x + 60/x = 30 + 4√15 + 60/√15
A = 38 + 4√15 + (60√15)/15 = 30 + 4√15 + 4√15
A = 38 + 8√15 ≈ 68.98 in²

Answer 3

Let L be the length of your text and W the width of your text. We also know L * W = 30

You have 2-inch top and bottom margins so the length of your page is (L + 4) and you have 1-inch left and right margins so the width of your page is (W + 2). The area of the sheet can be expressed as

A = (L + 4)(W + 2)
A = (L + 4)(30/L + 2)
A = 30 + 2L + 120/L +8
dA/dL = 2 - 120/L^2

Set your derivative equal to 0 and solve for L.

0 = 2 - 120/L^2
120/L^2 = 2
120 = 2L^2
60 = L^2
7.75 = L

So, the length of your text is 7.75 inches and the length of the page is 11.75 inches. The width of your text is W = 30/L = 30/7.75 = 3.87 inches so the width of your page is 5.87 inches.

The minimum area is 68.98 square inches.

Answer 4

a million. attempt picturing a sq. with 4 mini squares in each and every nook...the mini squares have a edge length x...as a consequence V=(3-2x)(3-2x)x V=4x^3-12x^2+9x dV/dx=12x^2-24x+9 or 4x^2-8x+3 serious whilst (8+-4)/8 so 12/8 or 4/8; 3/2 or a million/2 whilst x=a million/2, V=2; whilst x=3/2, V=0; as a consequence the respond is two cubic meters 2b. 2x+y=4 hundred y=4 hundred-2x A=xy A=400x-2x^2 dA/dx=4 hundred-4x serious whilst x=one hundred diagnosis making use of the by-product equation above: (0,one hundred)-beneficial (one hundred,4 hundred)-destructive hence, one hundred is a optimum dimensions would be one hundred x 200 for a community of 20000 sq. meters 2a. 50 is now the acceptable length the rectangle could be; so one hundred+y=4 hundred y=3 hundred 50 x 3 hundred = 15000 sq. meters