Calculus, optimization?

Question

Optimization. Find the dimensions of the rectangle of maximum area that can be enclosed by a three-sided fence (the fourth side is already closed by an existing wall) if the total length of the fence is 40 feet.

Answer 1

a + 2 b = 40

A = a * b = (40 - 2b) * b = -2 b^2 + 40 b

Find b by solving A'=0.

Answer 2

1) It is simple; ==> First decide the objective of the problem; here it is paper dimensions, so that its area will be minimum; then next comes under what constraints this has to be done? ==> Well, here On the selected paper, you are leaving some margins all over and arriving at the area for printing, which is given some fixed value. 2) Now, we are clear what we want; How to proceed for building mathematical equations, so that it can be solved for getting the end answer: 3) Here we are given the printing area as 50 sq in which is constant; hence let us start from this data; ==> Let the dimension of the printing area be x in (height wise) by y in (width wise) ==> Printing area = xy = 50; == y = 50/x -------(1) 4) There is a margin of 4 in each at top and bottom are provided; hence overall height of the paper is "x + 8" in; Similarly a margin of 2 in is provided on either sides; ==> Overall width = y + 4 in 5) Hence the area of the paper is = (x+8)(y+4) 6) Substituting for y from equation (1), A (x+8)(50/x + 4) = 50 + 400/x + 4x + 32 7) Hence the function to be minimized is: A = 82 + 4x + 400/x 8) Differentiating this, A' = 0 + 4 - 200/x^2 = 4 - 400/x^2 9) Equating A' = 0, x^2 = 100; ==> x = +/- 10 in 10) But a dimension cannot be negative, hence we consider only + 10; So x = 10 in and y = 5 in 11) However we need to verify,whether this is minimum or maximum; for which we will apply 2nd derivative test; So again differentiating, A'' = 800/x^3; at x = 10, A'' is > 0; hence it is minimum Thus we conclude for the printing the area to be 50 sq in, under the given constraints of margins, the outer size of the paper must be 18 in by 9 in; So outer area is = 162 sq in. Wish you are explained; have a nice time.

Answer 3

Let x be the length of each of the two congruent sides, and y be the length of the third side.
2x+y = 40
A = xy = x(40-2x) = f(x), where A stands for the area
Amax = f(10) = 200 ft^2 at x = 10 ft, y = 20 ft
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Reason: f(x) is a quadratic function, opening down. Therefore, the max area is at the vertex point where x = (0+20)/2 = 10.

Answer 4

first write the equation

area = l * w
perimeter = 2l + w = 40 (1 wall is already taken)
w = 40 - 2l

sub into area formula
area = l * (40 - 2l)
= 40l - 2l^2

take derivative and solve for 0

0 = 40 - 4l
10 = l

plug solved L into perimeter

40 = 2(10) + w
20 = w

final dimensions:
L = (10)2
W = 20

In english, 2 sides of 10 feet in length by 1 side of 20 feet in length

Answer 5

Suppose the width=x,length=40-2x,then y is the area
y=(40-2x)x
y=40x-2x^2
y'=-4x+40 When y'=0
x=10
As y''=-4,it has the maximum area,
so when x=10,y=(40-2*10)*10=200
Thus,width=10,length=20 at this time.

Answer 6

A = L * w
P = 2L + w = 40

w = 40 - 2L

Substitute the constraint into the objective:
A = L (40 - 2L)
= 40L - 2L^2

Take derivative:
A' = 40 - 4L

Set derivative equal to 0:
40 - 4L = 0
4L = 40
L = 10

Plug result back into constraint:
2(10) + w = 40
20 + w = 40
w = 20

So two sides are 10ft and one side is 20ft.