What is the maximum area possible? Explain.?
818428
perimeter = 2w + 2l = 440\nw + l = 220\n\nA = lw = w(220-w) = 220w - w^2\ndA/dw= 220-2w\nmaximum point occurs at dA/dw=0\n2w=220\nw=110\n\nMaximum area= 110(220-110) = 12100sq. ft\n\nI'm not quite sure whts the string of numbers about
117044
Area equals half the base times the height.  \nA=1/2bh.  Plug in and solve!
814915
Area = ½ ab SinC\n Area = ½(55)(29) Sin 61º\n Area = ½(1595) Sin 61º\n Area = 797∙5 Sin 61º\n Area = 797∙5(0∙874 619 707)\n Area = 697∙509 2165\n Area ≈ 697∙51 square units.
241144
to calculate the area of a 2-dimensional space you always multiply length x width, or in this case height\n\nC
794422
If h = height (and width), and L = length, 2h + L = 62 and so\nL = 62 - 2h\n\nVolume = lengthXwidthXheight = (62-2h)(h)(h) = 62h^2 - 2h^3\n\ndV/dh = 126h - 6h^2 \n\nMaximum occurs when this is zero\n0 = 6h(21 - h)\n\nSo maximum occurs when h = 21\n\nUse that to get the maximum volume
1100142
area or volume???\n\narea = 2 (3.4 * 3 + 3*2.2 + 2.2*3.4 )\n in square meters.\n\nvolm = 3.4*3*2.2\nin cubic meters.\n\n\nI only edited bcoz ever1 is giving wrong ans for area. Correct area is 48.56 sq mtr. I dont understand how it can b 6.6\n\nvolm = 22.44 cubic mtr
252742
I believe the bounds are where the numbers are deviated but still close. Say, in the problem, 11 cm is measured to the nearest centimeter. So, it may not be 11cm. It could be 10.6 cm, 11.3 cm, or even 10.51294803 cm. But they'll all be 11cm in the measurement. \n\nSo for the question, if it's asking for the minimum possible area:\nLet l be length and w be breadth. \n10.5 <= l < 11.5\n8.5 <= w < 9.5\nThe lower bound would be composed of the shortest length and breadth. \nA = 10.5(8.5) = 89.25 cm^2
87402
a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
778628
If you subtract the 0.3 sq. inches you will get the minimum.\n50.25 - 0.30 = 49.95 sq. inches\n\nAnd if you add the 0.3 sq. inches you will get the maximum.\n50.25 + 0.30 = 50.55 sq. inches
1117115
Limitations on file size would be my guess.
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~ ♥ ~
181180
The max you can have is 50x50, but you can have anything smaller than that.
236529
Since the sum of 2 consecutive sides is 50, try multiplying 2 such numbers until you get the biggest answer (since area = lengthxwidth)\n\n10x40 = 400\n20x30 = 600\n24x26 = 624\n25X25 = 625\n\nThat's the biggest area possible.
904847
The total area or the can is give by:\nArea of curved surface = 2π r h\nArea of top of can = π r²\nArea of bottom of can = π r²\n-----------------------------------------\nTotal area (A) of can = 2π r h + 2 π r²\n A = 2π r h + 2 π r²\n A - 2 π r² = 2π r h\n (A - 2 π r²) / 2π r = h\n h = (A / 2π r) - r\n\nIt's the best I can do for you based on the information given.
480877
i even tried trial and error, there is no solution to this equation. the result of equation can be maximum equla to 174 (98+76), i tried all the possible values by using trial n error , there is no solution to this. SORRY....!!!!
77962
Maximum: 1658500000 km \n\nMinimum: 1195500000 km
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72