calculus: question about finding greatest possible area?
919994
Find the 2nd derivative and find the max.
30455
i think you might mean what 2 dimensional figure has the greatest area in relation to its perimeter. in that case the circle has the greatest area of any shape with the same length perimiter.  so the answer is nothing.  if you take a rope and are asked to arange it into a closed shape with the most area, a perfect circle is the shape you'd use.
1019108
Do you have limits between which to calculate the area? If not you can solve for the values where f(x) = g(x)\nyou should get two values since g(x) is a second degree function.\nThe 2 values are your limits for integration.\nIntegrate each function with respect to dx between the limits and take the difference of the two. Since area is always positive the value should be the absolute value of the difference.\n\nI think that is enough hint. I won't do it for you. But here is what you should start with:\n\nsolve for x where f(x) = g(x), the two values for x are the two limits for your integration, for example x1 and x2.\nthen integrate\n  x1\n| ∫( f(x) - g(x) ) dx |\n  x2
246446
Y'all didn't answer the question! You just found the length of the sides. :)\n\nThe largest square that will fit inside has it's 4 corners touching points on the circle.  If the radius is 10 feet, then the diameter is 20 ft.  The diameter is also the diagonal of the square with side length x, and is therefore the hypotenuse of a 45-45-90 right triangle.\n\nx^2 + x^2 = 20^2\n2x^2 = 400\nx^2 = 200\n\nThat's as far as you have to go, since the area of the square is length (x) times width (x), or x^2\n\n~*~*~So the area of the largest square is 200 sq feet.~*~*~
907697
You need to find an equation first. try drawing the box out it may help. Draw a rectangle label one side 16 inches the other 30 then draw squares at each corner.\n\nLet x be the length of one of the square's sides\n\nVolume = length * width * height\n\n= (30-2x) * (16- 2x) * x\n\n=4x^(3) - 92x^(2) +480x \n\nTherfore  for largest volume: maximise V= 4x^(3) - 92x^(2) +480x \n\nTo maximise: first find values of x when the first derivative is equal 0:\n\nV ' = 12x^(2) -184x + 480 = 0\n\nx = 12 or 3 1/3\n\nNext verify which gives a maximum V value by either using first derivative test or second derivative test:\n\nWith first derivative test we choose values between intervals of x values for V'=0 and look at the trend:\n\nPositive values mean the function is increasing (/) and negative values means the function is deacreasing at that point (&#92;)\n\nx | 0 | 3 1/3 | 5 | 12 | 14                  \nV'| / .|... - .. | &#92;. |.. - ..| /\n\nTrend on either side of x= 3 1/3 :/&#92;\nTrend on either side of  x=12 : &#92;/\n\nTherefore max is at x= 3 1/3\n\nBy second derivative test :\nIf v"< 0 ie a negative number, V is a maximum\nIf v">0 ie positive number,V is a min\nIf V" = 0 it is a point of inflection\n\nV"= 24x-184\n\nx= 3 1/3,  V"= negative therefore max\nx= 12 ,    V"= positive therefore min\n\nHence the squares should be of lengths 3 1/3 inches to give the largest volume
892318
Take the right triangle to be arranged in the first quadrant.\n\nThe hypotenuse forms the line y = (-4/3)x + 4\n\nTake the dimensions of the rectangle to be x and y, subject to the above constraint.\n\nWe want to maximize A = xy.\n\nWe've solved for y above. This gives:\n\nA(x) = x[(-4/3)x + 4]\nA(x) = -(4/3)x^2 + 4x\n\nIf you're not taking calculus, graph this function and find the x that maximizes the function. It should be at the vertex of a parabola that points down - a clear indication that it is the maximum.\n\nIf you're taking calculus, find A'(x) and set it equal to 0.\n\nA'(x) = (-8/3)x + 4 = 0\n\nx = 4*(3/8) = 1.5\ny = (-4/3)* 1.5 + 4 = 2\n\nThe rectangle measures 1.5 x 2 cm, and has an area of 3 cm^2.
644833
if x is width, then 30-x is length \n(so that 2* (width_length)=60),\nand area is indeed 30x-x^2, which is maximized at x=15.\n\nGenerally, a rectangle with a given perimeter will have max area if it is square.
763428
QUESTION1 -- Profits = Revenue - Cost, so you want to come up with an equation for profits using R(x) and C(x), then use differentiation to determine your maximum.  Let P(x) = profits.  So:\n\nP(x) = R(x) - C(x) =\n50x - 0.5x² - 9x - 3 =\n-0.5x² + 41x - 3\n\nP '(x) = -x + 41\nP '(41) = 0  << so you have a min or max at x=41\n\nP ''(x) = -1\nP ''(41) = -1  << tells us x=41 is a max\n\nSo, answer for #1 = Maximum profit at x=41\n\nQUESTION 2 -- Let x = side of base, so L and W each = x since the base is a square.  The height is unknown, but we can solve for it since we know the volume.\n\nV = 57 ft³ = x²h,\nso h = 57/x²\n\nWe want to maximize surface area, so let's first get our formula (all in terms of x) for the surface area:\n\nS(x) = x² + 4(x*57/x²) = x² + 228/x\n\nMaximize same way we did above using differentiation:\n\nS '(x) = 2x -228/x²\n\nS '(x) = 0 when 2x = 228/x², i.e., when x³ = 114; i.e., when x = cube.rt(114) ft.\n\nSo the dimensions of your box are L = W = x = cu.rt.(114), and h =  57/x² = 57/114^(2/3) ft.\n\n~ ♥ ~
723835
You have the equation for the area in terms of theta. Take that formula and differentiate it with respect to theta. That shows how the area changes when theta is increased or decreased.\n\nSet that differentiated formula to equal zero.  This is because if the differential were greater than 0, then you could increase theta and get more area; if it were less than 0, then you could decrease theta and get more area.\n\nOnly if it's 0 is there no way to get more area.\n\nAfter you've set the differentiated formula to equal zero, you have an equation in theta which you can solve (and hence find theta).
592617
the area = (1/2)*(4+x)*(16-(1/4)*(x-4)^2)^(1/2).  you now take the derivative and set it equal to zero and solve for x.  this will give you the length of the forth side that would mazimize the area.  the derivative = (32+4x-x^2)/(2*(48+8x-x^2)^(1/2)) .  when you set this to zero and solve for x, you get that x=8....I got the derivative by applying the product rule of differentiation.  the math is a bit to long to type out in my answer
1151541
This is impossible:\nProof. Let the area of each identical square equal n^2 (so each side is n). Since the large square is has 3 of these pieces, its area is 3n^2. Thus, each of its side is sqrt(3)n. However, you can never divide the square to do this (since sqrt(3) is irrational).
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
480203
Area of a circle = pi*r^2\n\nr=radius os the circle
481212
the area under the curve f(x)!
786144
Multiply through the parentheses:\n\nA(x) = 3000x - 2x²\n\nNow, take the first derivative of this equation,set it equal to zero, and solve for x:\n\nA'(x) = 3000 - 4x = 0\n4x = 3000\nx = 750\n\nSo, the area of the space would be 750 x 1500 = 1,125,000 sq ft.\n\nYou integrate and set = 0 because you are essentially graphing the possible areas based on the length of one of the sides. When the tangent to the graph goes to zero, then the curve is at its maximum location.
936604
let the equal sides be x\n1/2x^2=area=20 given\nx^2=40\nx=rt40\nusing the pythagoras theorem\nc^2=a^2+b^2\nc=rt(a^2+b^2)\nthird side=rt(40+40)\n=rt80\n=4rt5 cm
907964
let l be the length and w be the width\n lw=111\n\nbut it has a margin so Area= (l-11cm)(w-4)\ndifferntiate\nbut we have 3 unknowns so we use the first ea lw=111\nl=111/w\n\nA=(111/w-11)(w-4)\n\ndA/dw= (111/w-11)+(w-4)(-111/w^2)=0\n\n111/w-444/w^2= 111/w-11\ncancel 111/w\n-444/w^2=-11\ncrossmultiply\nw^2=-444/-11\nw=6.4 cm\nl=111/6.4=17.3