Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use t

Question

John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation.

Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex form to find the maximum area.

Answer 1

A square encloses the largest rectangular area.

With 300 feet of lumber, you can enclose a square with 4 sides each 75 feet.

So the patio would be 75 feet x 75 feet (5625 sq. ft.)

If you need proof that this area is the largest, write the equation for the area and perimeter of the rectangle:

W x L = area
2 (W + L) = perimeter

perimeter is fixed at 300 feet.
2 (W + L) = 300

Solve for W:
W + L = 150
W = 150 - L

Now plug W into the first equation:
f(L) = L (150 - L)
f(L) = -L² + 150L

If you know calculus you can take the derivative and set to 0, this will be the maximum:
f'(L) = -2L + 150 = 0
150 = 2L
L = 75

And solve for W:
W = 150 - L
W = 150 - 75
W = 75

So the maximum is L = 75 feet, W = 75 feet (a square).

If you don't know calculus, you can just graph the area against length:
f(L) = -L² + 150L

Make L = x, and graph the function:
y = -x² + 150x

When you graph it, you'll find that it is an upside down parabola that crosses the x-axis at (0,0) and (150,0). Because the parabola is symmetric, the maximum is halfway between 0 and 150 (75). So the maximum point is at x = 75, y = 5625. Any other lengths and you'll have a smaller area.

This proves the maximum area is a square with sides of 75 feet by 75 feet.

Answer 2

P = 2L + 2W; A=L x W

We know P = 300, so substituting:

300 = 2L + 2W or 150 = L + W

Solve for one variable:

L = 150 - W

Now substitute this into the area formula A = L x W

A = (150 - W) x W or

-W^2 + 150W = A

Now we use some calculus to calculate the maximums:

F(A) = -W^2 + 150W find the derivative:

F'(A) = -2W +150

Now solve:

0 = -2W + 150
2W = 150
W = 75

From the original solved equation:

L = 150 - W
L = 75


With 300 feet of lumber, the design to maximize area would be a rectangle with dimensions of 75 x 75 (a square!)

Answer 3

ok, all of us comprehend the edge of the rectangle is = 2 x length + 2 x width so 2l + 2w = 3 hundred and area = l x w so which you would be able to desire to locate the area of the rectangle with the records you have have been given now, we are able to make certain that w = a hundred and fifty - l hence, the area (A) = l(a hundred and fifty-l) = 150l - l^2 the area of this rectangle ameliorations as length and width exchange. If the area is graphed against the size, you will get a curve popping out, shaped like a parabola, with the equation y = 150l - l^2. be conscious the coefficient of l^2 is detrimental, which makes the parabola concave down. Concave down parabolas have a optimal value, earlier they get smaller lower back. to locate the vertex, the formulation is -b/2a = -a hundred and fifty / 2 (-a million) = seventy 5 because of the fact of this the axis of symmetry, the place the parabola's y value is optimal, is alongside x=seventy 5, which for this reason, ability l=seventy 5. by subbing l=seventy 5 into y= 150l - l^2, we finally end up with y = a hundred and fifty x seventy 5 - seventy 5^2 = 5625 sq. feet which by the way is seventy 5^2. This shows that the optimal area would have been discovered by dividing the 300 into 4 equivalent factors, and making a sq., because of the fact the sq.'s area is often the perfect while the fringe is fixed. besides, that final bit became into extra information, that's no longer possibly area of the respond. desire the different stuff enables :)

Answer 4

Well, I don't know anything about the vertex method, but you start with two equations:

1) 300=2W+2L
2) A=W*L

first you solve 1) for either W or L: W=150-L, then sub into 2): A=L*(150-L). Next, take the derivative of A with respect to L and set it equal to zero: 0=150-2L -> L=75. Plug this back into 1) and you see that W=75 also. This proves what we all know, that a square has the maximum area/perimeter ratio for a rectangle.

Answer 5

If he made the cover round, he would shade more area so when the sun move he would still be in the shade