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If a farmer has 100 meters of fence what is the maximum area if he has a barn?
Wiki User
∙ 14y ago
The maximum area is 625 m². The maximum area can be found using a system of equations with two unknowns. By expressing the Area equation with only a single variable, the area is maximized wherever the first derivative of the equation is equal to zero. We find that the area is maximized when the two opposite sides of the barn are set to 25 m. Therefore, every side mesures 25 m and the maximum area is 625 m².
Wiki User
∙ 14y ago
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What is the maximum area of a shape with a perimeter of 1000 m?
This might be a bit of a cheap answer, but: Of all shapes, circles have the highest area/perimeter ratio; that is, they have a maximum area for a minimum perimeter. 1000 = circumference = 2 pi r r = about 159.155 m area = pi (r squared) = about 79577.538 sq. m using pi = 3.14159
How many 8 x 4 bricks to cover an area of 14 x 14?
8x4=32. 14x14=196. 196/32=6.125. You can use a maximum of 6 bricks, which will only cover 192 of the 196 square units.
Find the maximum area of a rectangle inscribed in a circle of radius 12 in?
The length and width of the rectangle (and thus it's area) will be defined by the circle. Specifically, the diagonal of that rectangle will always be equal to the diameter of the circle. To find the maximum area then, one can simply define the rectangle with that relationship, take the derivative of that function, and find out where that comes to a value of zero. The area will be peaked at that point: a = lw d2 = l2 + w2 ∴ l = (d2 - w2)1/2 Remember that the radius is 12, so the diameter is 24, and things can be simplified by plugging that into the equation at this point: l = (576 - w2)1/2 ∴ a = (576 - w2)1/2 w The next step is to take this equation for area, and find it's rate of change with respect to width: ∴ da/dw = (576 - w2)1/2 + 1/2 * (576 - w2)-1/2 * -2w * w ∴ da/dw = (576 - w2)1/2 - (576 - w2)-1/2 * w2 Now let that value equal zero: 0 = (576 - w2)1/2 - (576 - w2)-1/2 * w2 ∴ (576 - w2)1/2 = (576 - w2)-1/2 * w2 ∴ 576 - w2 = w2 ∴ w2 = 576 / 2 ∴ w2 = 288 ∴ w ≈ 16.97 And one can work out the corresponding height: d2 = l2 + w2 ∴ 242 = l2 + 288 ∴ l2 = 576 - 288 ∴ l2 = 288 ∴ l ≈ 16.97 Meaning that the optimum rectangle is a perfect square. To be completely thorough, it should be confirmed that the area found was a maximum and not a minimum. This can be done easily enough by taking a slightly lesser width and a slightly greater width, and seeing how their areas compare: a = (576 - w2)1/2 w Let w = 16 ∴ a = (576 - (16)2)1/2 * 16 ∴ a = 3201/2 * 16 ∴ a = 51/2 * 128 ∴ a ≈ 286.22 Let w = 18 a = (576 - 182)1/2 * 18 ∴ a = 2521/2 * 18 ∴ a = 71/2 * 108 ∴ a ≈ 285.74 Both of these values are less than the area surrounded by the square, so the square's area is indeed a maximum and not a minimum.
Consider an isosceles triangle whose two equal sides are of length 4 What is the largest possible area for such a triangle?
An isosceles triangle with side length 4 has an altitude x. By the Pythagorean theorem, the base of the triangle is 2*SQRT(16-x2). The area of the triangle is 1/2 base times height, so A=x*(16-x2)1/2. the derivative, dA/dx=(16-x2)1/2 - x2/(16-x2)1/2. This is found with the product rule and chain rule. This shows the rate which the area of the triangle changes with respect to the altitute. At the x value of the maximum, the area will have stopped increasing and begun to decrease, so the rate of increase wil be zero. We just need to solve for x. (16-x2)1/2 - x2/(16-x2)1/2=0 (16-x2)1/2=x2/(16-x2)1/2 (16-x2)=x2 16=2x2 8=x2 SQRT(8)=x. Now we can solve the original equation for the maximum are. SQRT(8)*SQRT(16-8) SQRT(8)*SQRT(8)=8 So 8 is the largest possible area.
What formula for throat area?
You cant find the area of the throat because the throat is 3-D and the area is only for 2-D measurements
How much maximum area can 100 ft of fence hold in a rectangle?
The maximum area possible using 100 feet of fencing is that inside a circular fence. But, since you asked about a rectangular fence: a 25x25 fence would hold 625 square feet. a 10x40 fence would hold 400 square feet. a 5x45 fence would hold 225 square feet. As you can see, the square is the rectangle that encompasses the largest area, as it is closest to a circle.
What is the maximum area you can enclose with 100ft of fence for a rectangular garden?
625 sq feet.
what- The farmer has a field where the fences are also perpendicular bisectors. What is the area of the field in square meters?
8,536 square meters
Which weapon has the greater maximum effective range M4A1 or the M16A3?
the M16A3 has a maximum effective range of 550 meters for a target and 800 meters for area, the M14A1 has a maximum effective range of 360 meters and 500 meters respectively.
What is the maximum range of of the M4A1?
500 meters at a point target, 800 meters at an area target.
What is the area of a fence that measures 60 m X 60 m?
3600 square meters
A farmer wants to build a rectangular garden with river as one side if he has 220m of fencing materials What is the length of side of the fence perpendicular to the river to enclose the biggest area?
55 meters. This makes the length parallel to river 110 meters, total area = 55 x 110 =6050 sq meters. I did this using differential calculus. If you don't know this, you can try other combinations. Put short side equal to 50 or 60 meters and work out area, it will be less!
A farmer has a rectangular garden plot surrounded by 200ft of fence. Find the length and width of the garden if its area is 2400ft2.?
12 feet
Farmer john has 100 yards of fence to put around his garden what shape should he make to maximize the gardens area?
rectangle
The Phillips family wants to fence their backyard They know the yard has a perimeter of 24 meters and an area of 32 square meters What is the yards length and width?
8 meters and 4 meters
What is the maximum effective range of a M240B on a area target?
3,725 meters
What is the effective range to a point area target and maximun effective range of the M249 squad automatic weapon with the short barrel?
The maximum effective range of a point target on the M249 is 800 meters. The maximum effective range of an area target is 1000 meters. The maximum range is about 3600 meters.
