the area of a rectangle is 18cm whats its greatest possible area?  another ?   ummm...the area of a rectangle is 36cm2 whats its least possible perimeter?  SMART PEOPLE HELP ME PLZ!!!! I WILL LOVE YOU FOR EVER!!! Email me if you want 2 talk
136600
For the first, if one side is x, then another side would be 9-x (so that they would add up to 18).  The area is then\n\nA = x(9-x)\n= 9x - x^2\n\nTo find the maximum area, take the first derivative and set it to 0:\n\nA' = 9 - 2x\n0 = 9 - 2x\n2x = 9\nx = 9/2\n\nSince the parabola opens down, the maximum area happens when x = 9/2.\n\nSo if x = 9/2, the other side is 9 - 9/2, or 9/2 - it turns out to be a perfect square!  And the area is then (9/2)^2, or 81/4.\n\nFor the second, you already have an area and you need to find the perimeter.  So, lw = 36, and we're trying to find the smallest 2l + 2w.  Well, l = 36/w, so the perimeter can be expressed as\n\nP = 2l + 2w\n= 2(36/w) + 2w\n= 72/w + 2w\n\nWe can take a derivative and set it to 0 to find the minimum value here:\n\nP' = -72/w^2 + 2\n0 = -72/w^2 + 2\n72/w^2 = 2\nw^2 = 36\nw = 6\n\nSo if w = 6, then l = 6 as well, and the perimeter would then be 24.
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.
470542
The smallest possible perimeter will be when the side lengths are nearest to the root of 72 (or closest together)\n\nsq(72) = 8.48....\n\n8 and 9 are nearest.  The perimeter is 2(8) + 2(9) = 34\n\nThe largest possible perimeter is when the values are farthest apart.  In this case, that would be 72 and 1.  The perimeter is 2(72) + 2(1) = 146.\n\n146-34 = 112.  That is your answer.
72736
I'm not sure if this is what you're asking, but if...\nd = length\nd-3 = width\n\nthen the area is d(d-3), a.k.a. d^2 - 3d .\n\nHope this helps!
426057
For this problem,\n\nArea of rectangle = length*width.  The length is defined by 2x because the two bottom corners are and 'x' distance from the origin.  The width is determine by the curve 16-x^2.  This is the constraint we need to use.  In other words:\n\nArea = (2x)(y)\n\nSince y = 16-x^2, substitute:\n\nArea = (2x)(16-x^2) = 32x-2x^3\n\nTo maximize area, take the derivative and set equal to zero:\n\nA' = 32-6x^2 = 0, solving for x, you get = +4*sqrt(3)/3, -4*sqrt(3)/3.  We need to take positive numbers because we are dealing with area, which cannot be a negative number.\n\nNow plug this extreme value into the area.  We know that \n\nlength = 2x\nwidth = 16-x^2, now we can find the dimensions of the rectangle.\n\nLength = 8*sqrt(3)/3\nWidth = 16-(4*sqrt(3)/3)^2 = 16-(16/3) = 32/3\n\nThese are your dimensions\n---------------\n\nHope this helps
106822
Perimeter\ntwo sides are 8 and the other two sides are 16. \n\n8 times 2 is 16 and 16 times 2 is 32. \n\n16+32= 48(unit) \n\n                Area\njust multiply them:) \n\n8 times 16= 128(unit) squared\n\nDoes that help?
638700
Because you have a fixed height, h, of 30m, and the Volume  of a triangular prism is given by: V = h*A where A is the area of the triangular base, your problem reduces to maximizing A, given the perimeter of A = 210m\nAn equilateral will enclose the maximum area(see link below) So the length of each side will be 210/3 = 70 m. The area of an equilateral triangle =  S^2 x sqrt(3)/4 where S = side length. So, in your case the area = 70^2*sqrt(3)/4 = 2,121.18 m^2 and the maximum volume for your prism is 30m*2121.18 m^2 = 63,652.86 m^3\n\nFor a polygon, the general rule is: the 'closer' the shape of a polygon is to a circle, the greater the area of the interior of the polygon. This implies that a equilateral triangle, a square(see below), a regular pentagon; and in general a regular n-gon will enclose the most area. \n\nConsider a square with side length X, then the area of the square = X^2 and the perimeter is 4X. Let's keep the perimeter fixed at 4X and see what happens to the area if we let one pair of opposing sides of the rectangle be X-y and the other two be X+y (Note the perimeter will still be 4X). However, the area is now: (X+y)*(X-y) = X^2 - y^2. We see that if y>0 then area decreases, while the perimeter stays the same. We see that a square encloses the maximum area for a fixed perimeter. \n\nHowever, if you allow the height, h, to vary as well then you have 3D maximization problem; but the solution is not difficult in this case either. In general, right-angle prisms with regular polygon bases will have a maximum volume for a fixed perimeter or surface area\n\nhttp://mathforum.org/library/drmath/view/53789.html\n\nhttp://mathworld.wolfram.com/Prism.html\n\nhttp://mathworld.wolfram.com/EquilateralTriangle.html
269850
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
532445
area: l*w = 256\n-> 4w*w = 256\n->w(sqaure) = 64\n->w=8 (width)\n->l = 4w = 4*8 = 32(length). \n\nThere you go.
287960
Assume that the length is y cm.  Then the width is (y-3) cm.\nTo find the area, multiply length by width : y x (y-3).  But this is equal to 54 cm^2.\n\nSo, y x (y-3) = 54\n    y^2 - 3y  = 54\n    y^2 - 3y - 54 = 0\n\nSolving for y (using any method at your disposal (formula, completing the square, factoring), y = 9 or -6.  Obviously, -6 cannot be a side.  That leaves us with the length being 9 cm.  If the width is 3 cm less, that will be 6 cm.\n\nYou did ask for it to be explained.
465391
A = lb = 98 in² (l = length and b = width, both in inches)\nSo b = 98/l\nWorking area W = (l - 8)(b - 4)\n= lb - 4l - 8b + 32\n= 98 - 4l - 8b + 32\n= 130 - 4l - 8b\n= 130 - 4l - 784/l\n\ndW/dl = - 4 + 784/l²\n= 0 for stationary points\nSo 4 = 784/l²\nie 4l² = 784\nThus l² = 196\nl = √196 (ignoring negative result)\n= 14 inches\nd²W/dl² = - 2 * 784/l³ < 0 when l = 14 so area is minimised when l = 14\n\nWhen l = 14 b = 7 (lb = 98)\nSo the dimensions to minimise paper used is 14 inches X 7 inches
511444
I wish I had Adobe Acrobat so I could e-mail you the lengths of the squares because explaining this without the diagram will be confusing, but I will try.\n\nThe important thing is to remember that the problem stated that this entire figure is a QUADRILATERAL THAT IS NOT DRAWN TO SCALE. So you can't assume that this is a square.\n\nLet's start with the side marked 9cm and 8cm (the bottom side). The small square in between the two is 1cm around (the difference between the 9 and 8 cm squares). Because of that 1 cm square, the larger square on top of the 9cm square is 10cm around (the side of the 9cm square + the side of the 1cm square). So on the left hand side of the quadrilateral, you now have a side of 9cm and 10cm.\n\nWe established the 1 cm square. The square on top of the 8 cm square is 7cm in length (8-1). So the last square on the bottom of the quadrilateral is 15cm (8cm+7cm). So you now have the total length of the bottom length of the quadrilateral (9+8+15)== 32cm. You also now have a partial length of the right hand side of the quadrilateral.\n\nThe square on top of the 7cm square is 4cm around. Subtract the 1cm side from the 7cm side (6cm), then subtract it from the 10cm side...(4cm). Now that you know that the square is 4cm around, then the remaining length on the 7cm side of the square beneath the 4cm square is 3cm.  (7-4). We already established the 15cm square. Add the 3cm length to 15cm to get 18cm and the remaining side of the right part of the quadrilateral, 33cm (18 +15).\n\nNow that you know 18cm square, subtract the 4cm square from it to leave you a length of 14cm. That leaves you with the remaining side of the left hand of the quadrilateral (10+9+14) ==33cm. It also leaves you with the remaining upper side of the quadrilateral, 14cm to give you (18+14)cm == 32cm\n\nSo you have 32+32+33+33==130cm, the perimeter of the quadrilateral, which is a rectangle.\n\nThis is just one way to solve this problem. There are others. For example, once you deduced the 4cm square, you could have added that side to the 10cm side to give you 14cm.
103896
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
608830
The maximum area would be a circle. If it has to be a rectangle a square would be the highest area. So divide 50 by 4 and you get 12.5 and that will be the largest. It is a square 12.5 feet per side.
525950
I hope you made a start.  Since you're asked the original side measure, let that equal x ft.  Then read the question again and notice that the length of the increased garden is x + 3, and the width is ...?  [I'm sure you know].\nArea = length * width, and we're told area is 72, so you have an equation in the form\n(x + 3)(x + ...) = 72\n\nExpand the left side to get 3 terms, then subtract 72 from both sides and get \n\nx^2 +  ...x - ..... = 0, which I hope you can solve, but I really don't think I should do it for you.  If you need more help, email the bloke who answered your previous question.\n\n-- or me at h_chalker@yahoo.com.au
457057
the question is not clear enough\n\ni think the lenght of cross section is the lenght of diagonal note it by d\nso let one side be l, the other side will be sqrt(d^2-l^2)\nThen the area is A(l) = l*sqrt(d^2-l^2).\nthe derivative of this is ...\nA'(l) = (d^2-2l^2)/(sqrt(d^2-L^2)) which is zero when d = sqrt(2) l, which happens only when the rectangle is a square. It remains to show that is a minimum( not a maximum). I let you do this...
384085
w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm