A settling pond @ a sewage treatment facility is rectangular. The length of the pond is 15ft more than 4 times its width w. What is the area of the pond?
119582
l = 4w + 15\n\nArea = l * w\nArea = (4w + 15) * w,  then\n\nArea = 4w2 + 15w
567721
Area=width*length=12*9=108 i.e \nC
215020
The area of the swimming pool is 15 times 30 which is 450.\nThe area of the pool and the border is\n450+424\n= 874\nLe the width of the border be x.\nSince it is uniform, the width of the pool with the border is 2x+15 and the width is 2x+30\nThe area of the pool and border is\n(2x+15)(2x+30)\nWe know the area is 874\nThus,\n(2x+15)(2x+30) = 874\nFOIL left side\n4x^2 + 90x + 450 = 874  Subtract 874 from each side\n4x^2 + 90x - 424 = 0\nDivide bot sides by 2\n2x^2 + 45x - 212 = 0\nFactor\n(x-4)(2x+53) = 0\nx = 4 or x = -53/2\nx cannot be negative.\nThus, x = 4\nThe length of the border is 4\n\nNow a cement block is 2ft by 2ft.  The area of the block is 2 times 2 which is 4\nTake the area of the cement border and divide by 4.  This will give you how many blocks you need.\n424/4 = 108\nThus, you need 108 blocks.\nEach one costs 1.24\nTotal cost = 1.24(108) = 133.92\nAdding 7% sales tax gives you\n133.92 + 133.92(.07)\n= 143.29
436129
perimeter=4 times the side\n=4*64=256 foot
599704
Since one litre = 1000cm^3,   We can construct a cylindrical container with surface area S = 2pir^2 +2pirh.(I am using pi since I don' t have the greek symbol.)  We wish to minimize S subject to the constraint that V = pir^2h = 1.  Now, rewrite the equation for S as S = 2pir^2 + 2pir(1/pir^2) = 2pir^2 + 2/r).  Now, differentiate S wrt r and you get ds/dr = 4pir - 2/r^2 = 0 => r = 1/pir^2.  Now , use h = 1/2pir^2 to solve for h.\n\nFor the rectangular boxes, the problem is impossible without some relation between the three sides..\n\nLet me know if you don't understand.
273178
Let the dimensions of the rectangle be 3x and 5x.\n\n3x * 5x = 135\n15 x^2 = 135\nx^2 = 135/15\nx^2 = 9\nx = 3 or -3\n\nAs x is a length, -3 is not applicable.\n\nSo the dimensions are 3(3)=9 by 5(3)=15, i.e. 9 by 15.
559013
Let the breadth of each red rectangle be x\n\nSo, area of red portion = 4x + x(3-x)\n\nSo, area of white portion is (4-x)(3-x)\n\nSo, 4x+x(3-x) = (4-x)(3-x)\nor 4x + 3x -3x^2 = x^2 -7x+12\nor 4x^2-14x+12 = 0\nor 2x^2-7x+6 = 0\nor 2x^2-6x-x+6=0\nor(2x-1)(x-3)=0\n\nx=1/2 or 0.5 foot is the reasonable answer.
33674
1.)The area of a carpet is 308sq.m. If its length is 2m less and its width 6m more, the carpetwill be square. Find the dimension of the carpet...\n\nSo we can see the length of the carpet is 8m longer than the width\n\nL B = Area\nL (L-8) = 308\nL^2 - 8L - 308 = 0\n(L-22)(L+14) = 0\n\ntherefore L=22m or L= -14m\n\nNow a carpet that was -14m long wouldn't be very useful, so we discard that result as not representing the real world.\n\nL = 22 ; B = 308/L = 14m\n\n2.) Gerry wants to fence a rectangular field whose area is 1200sq. m. He has only 100m of fencing so he decided to fence only the three sides of the rectangle letting the wall be the fourth side. How wide should the rectanglebe??? \n\nLet a and b  be the length of sides of the field\na.b = 1200       ------(1)\n2a+b = 100       ------(2)\nb = 100 - 2a     ------(2a)\n(100-2a)(a) = 1200     (substituting into eqn 1)\n2a^2-100a+1200 = 0     (expanding and collecting terms)\na^2-50a+600 = 0        (divide by 2)\n(a-20)(a-30) = 0\na = 20 or a =30\nb = 100-2a\nif a = 20, b = 60      (Check L=20+20+60,A=20*60 correct)\nif a = 30, b = 40      (Check L=30+30+40, A=30*40 correct)\n\nSo two feasible answers.
395797
let,Pump A and pump B can themselves empty the pond in x and y hours respectively\nTherefor in 1 hour pump A can empty 1/x part of the pone,and\n               in 1 hour pump B can empty 1/y part of the pond.\nBy the problem\n10/x +6/y=1.........(eqn. 1)\n6/x+12/y=1...........(eqn.2)\nmultiplying equn 1 by 2 and subtracting it from 2,we get\n-14/x=-1\n=>-x=-14=>x=14\nPutting the value of x in eqn 1 we get\n10/14+6/y=1=>6/y=4/14   =>4y=84    =>y=21\nThe pump A can empty it in 14 hours and pump B can empty it in 21 hours
461481
long sides are 24ft and short sides are 9ft.\n\nArea = width x lenth\n216 = 24 x 9\n\nDimensions 24+24+9+9 = 66
13832
Draw a rectangle to represent the lot. Lable the long side 240 and the short side 180.\nNow draw a smaller rectangle inside it so that it is the same distance from the large rectangle.\nLabel the distance between them (the lawn) as x.\n\nYou know the area of the small rectangle is its height times its width. If the height of the large rectangle is 180, then the height of the small rectangle will be 180 minus two strips of lawn or (180 - 2x). The width of the small rectangle will be 240 minus two strips of lawn or (240 - 2x).\nThe area of the small rectangle will be Width x Height or (180-2x)(240-2x).\n\nSince the area of the lawn is equal to the area of the factory, the area of the factory has to be half the area of the whole lot or 1/2x180x240 = 21600, so\n\n(180-2x)(240-2x) = 21600\n\nFactor out some 2's to make the number smaller and we have\n2(90-x)2(120-x) = 21600\n(90-x)(120-x) = 5400\nor\nx^2 - 210x + 10800 - 5400 = 0\nx^2 - 210x + 5400 = 0\nfacto\n(x-180)(x-30) = 0 solutions 180 and 30.\nSince 180 is not a viable solution the only possible one is 30.
55421
L x W = Area\n\nCheater
432041
I hope you meant to type "what is the width of the rectangle."  If there really is a triangle in this problem I'll need more info.\n\nAssuming it was the rectangle you meant to ask about, the first thing to do with a word problem is write equations for the data.  We'll call the width W, the length L and the perimeter P.  We're told that one side is 15 cm longer than the other.  We aren't told which is which, so I'll say it's longer than it is wide.  Therefore L = W + 15.  We're also told that the perimeter is 94 cm.  The perimeter is the sum of the lengths of all four sides, so it's 2L + 2W.\n\nSo our equations are:\nL = W + 15\n2L + 2W = 94.\n\nNow, substitute the first equation into the second:  2(W + 15) + 2W = 94.\n\nDistribute the 2:  2W + 30 + 2W = 94.\n\nSubtract 30 from each side: 2W + 2W = 64.\n\nCombine the Ws: 4W = 64\n\nDivide by 4: W = 16.\n\nNow go back to the first equation:  L = W + 15.  So L = 16 + 15 = 31.\n\nThe rectangle is 16x31 cm.  I don't know which is the length and which is the width, I just arbitrarily picked, so the answer could be either 16 or 31.  Hopefully you have additional information that will allow you to decide which it is.
34497
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.
454992
area of the rectangle=bh\n=6*10=60 sq.unit\narea of the traingle=1/2bh\n=1/2*2*10=10 sq.units\nratio=10/60=1/6
347523
Ok... the first thing to realize is that you are finding the maximum area with the least amount of materials.  To do that you should put the shed at a corner of the rectange, that way two sides of the shed act as part of the fence.  Assume that the shed is not "usable area" for the shep pen.  Accordingly, you need to solve the problem twice, once with using the shed as part of one side, and once using the shed as a corner.\n\nOk, the first thing you need to do is show that the maximum area of a rectangle occurs when the lenght and width are equal.  That is the only part that needs any type of equation analysis, and it really doesn't require calculus.\n\nA=L*W\nAssuming that the width is at most as long as the L, then the equation goes to:\nA=L*(x*L), where x a percentage, ranging from 0 to 1.\nClearly, the area (A) will be maximum when x=1.  That shows that in your problem, the pen will have the maximum area when the length and width of the rectangles are equal.\n\nSo, you know that the fence is 160m., and you have 4 sides, s1, s2, s3 and s4.\nSo:\n160m=s1+s2+s3+s4\n\nNow, if you use the shed as part of one side, namely s4, then the perimeter equation for the shed is:\n180m=s1+s2+s3+s4+20m\n\nSince the maximum area occurs when the lenght and width are equal:\ns1=s2=s3=s4-20m, so substituting s2, s3 and s4 for s1 in the permiter equation, you get:\n4s1+20m=180m\nSolve for s1,\ns1=45m (call this the length)\ns4=45-20=25m\nin this case, the area will be S1*S1, since the lenght and width of the pen will be equal.  \nSo the area is 45*45=2025 sq. m.\n\nNow, put the shed in a corner of the sheep pen.\nSo in our example, we'll make the shed part of s1 and s4.\n\nso the primeter equation will be:\ns1+20m+s2+s3+s4+20m=200m\nNow, s1=s2-20, s3=s2 and s4=s2-20\nSo substituting for s1, s2 and s4, you get:\ns2-20m+s2+s3+s4-20m+40m=200m\nsolve for s2,\ns2=50m=s3 and s1=30m (s2-20m)\nSo max area is 50m*50m=2500 sq. m\nBut, now you need to reduce the area by the area of the shed (400 sq. m), so final usable area is:\n2100 sq. m which is more than using the shed as part of only 1 side.
416583
The 400 feet of fencing forms the perimeter P of the rectangular area.  The perimeter of a rectangle is P = 2*W + 2*L.  The area of a rectangle is A = L*W.  Take either the W or L from the perimeter equation and solve for it in terms of the other.  For example. 2*W = P - 2*L, or W = (P - 2*L)/2.  Put this expression for W into the area formula\n\nA = L*(P - 2*L)/2\n\nA = P*L/2 - L^2\n\nTo maximize area, take the derivative of A with respect to L to get\n\nA' = P/2 - 2*L  and set it equal to zero\n\nP/2 - 2*L = 0, or  L = P/4;  This gives you the length of the area.  You can get the width from the perimiter formula W = (P-2*L)/2;\nPutting in L = P/4 gives W = (P - P/2)/2 = P*(1-.5)/2 = P/4.\n\nNotice that L = W = P/4; in other words the maximum area rectangle is a square.