Help I need to finish this prob. What is the greatest possible difference between the areas of two rectangles?
89149
Take the two extremes of the length of the sides.  The shortest a side can be is 1cm.  With a length of 1cm, the width would be 71 cm (1+1+71+71=144).  This is an area of 71 (1 x 71=71), which is the smallest possible.  The largest possible area is of a square with perimeter 144, which is each side being 36cm.  This is an area of 1296.  So, the difference is 1296-71=1225.
736325
Area of a rectangle is length XWidth. So if you have the area of a rectangular field or courtyard and either length or breadth/width,just divide the area by length to get the width and similarly just divide the area by width/breadth to get the length it is very easy\nWe know that Length XBreadth=area of the rectangular yard.\nTherefore,length=360.15/15\n=24.01m
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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.
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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...
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a ) you effectively have a perimeter of 180m because 160 plus 20. \n\n so you take 180/4 because it;s retangle.  so your area would be 45 x 45 = 2025 m^2
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First pick what you want your area to be.\nLet A = 100.\nThe first one is easy, just let it be a square with all side lengths of 10\nNow pick two other numbers that when you multiply them you get 100.\n4 times 25 = 100\nA rectangle with width 4 and length 25.\nKeep picking more\n2*50 = 100\nwidth = 2 and length = 50\nYou can find more.\nAlso, you don't have to use 100, just pick any number you want for the area.
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All rectangles have equal diagonals.\nAnd a square is one example of a rectangle.
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I am not sure if I understand your question correctly, or else your question is not complete.\nFor example, a usual question of this sort is to find the maximum area of a rectangle for a given perimeter, the classical fence problem, as follows:\n\nYou are given a specific length of fence L to make en enclosure of maximum area.  What would be the dimensions of the rectangle.\n\nHere we take the length of one side as x, then the other side will be (L-2x)/2.  The area is thus A(x)=x(L-2x)/2.\n\nTo obtain the value of x for a maximum area, we differentiate A with respect to x, \ndA(x)/dx = d(Lx/2-x^2)=L/2-2x\nfor maximum area, dA(x)/dx=0, so we equate\nL/2-2x=0\nand solving for x gives x=L/4.\n\nThat means that the maximum area rectangle for a given perimeter is a square!\n\nHope that helps with your project.\nPost if you need other information, but please post details, especially how far you are with your project, so we can help you better.
46001
you need to be more specific...\nbut technicaly the perimeter is the sum of all the sides...
222548
Your rectangle (in feet) comes out to 8.2' x 19.7'.  Each row needs 8 tiles plus one strip of 1x6 tiles (if your tile sheet is 2" tiles in a 6x6 configuration).  You will have 19 full rows of tile in the other direction, plus 1/2 sheets of 3x6 tiles.  All in all, you will use 152 FULL tiles, plus 1.3 sheets for the 1x6 strips and 9.5 sheets for the 3x6 sheets.  All in all, you will have 12 tiles left over (2 1x6 strips) if everything lays out properly.\n\nTotal number of tile sheets needed: 163\nTotal left over: 12 2" tiles.
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1.rectangle\n\n2.trapezoid
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Once again you will see that this is a rectangle.\nA = length times width\nA = 7*2\nA = 14
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Graph them both.  Notice that, after graphing the one, you only need to turn your "page" 90 deg counterclockwise and you'll get a mirror image of f^-1(x).  \n\nYou'll also notice that each has exactly one rectangular area that the other does not.  The first, f(x), is a "piano"-shape, formed by a "pseudo triangle" and a rectangle, the latter of which measures 8x5, because 8=12-4.  Now take that rectangle away, and look from the point of view of f^-1(x).  This "triangle" lacks a rectangle 4x4 in dimensions, where the second 4 is the height, from f^-1 (9) = 4, and the first 4 is from 4=9-5.\n\nThus, we subtract 8x5=40, and add 4x4=16, to get our answer (which I'm actually too lazy to finish giving; it's 24 less than the first area).
800206
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.
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So you've got an eight-sided figure and each side is 72-feet.  If you draw horizontal and vertical lines from each vertex (the angles where sides meet) to the other side of the octagon, you'll see that the octagon can be split up into one large square (in the center), four rectangles (up, down, left and right), and four triangles (in the corners).  You following me?  Get out a pencil and get to it!  Okay.  So what are lengths of your square's sides?  You'll note that the square has 72-foot sides, just like the original octagon.  Since you know how to find the area of the square, you're on your way.\n\nNext.  you have four rectangles, each with a length of 72-feet, but the short end - the width, how do we figure out the width of the rectangle?  Okay.  Look at your triangles.  Each triangle has one side of 72 feet - the side which is the outside of the octagon - and then two equal sides - it's an isosceles triangle with a 90-degree angle.\n\nOkay, time to call on pythagoras.  Pythagorean theorem states that a^2 + b^2 = c^2.  What that means is this: If you take the length of sides a and b (the ones we don't know) and square them, and add them together, the answer will be 72-squared (side C).  Too bad I'm not there with you to make this part easier to understand.  We know that a and b are equal in this case because our triangle is isosceles.  Therefore, 2x^2 is equal to 72^2.  This gives us a messy equation wherein our answer is, x is ROUGHLY equal to 51 feet.\n\nSo.  You have a square with sides 72 feet long.  You have 4 rectangles each with length 72 and width 51.  You have 4 triangles each with base 51 feet and height 72 feet.  You know that squares and rectangles have an area equal to length times width, and triangles have an area equal to one-half base times height.\n\nCalculate the area for your 9 figures and add them together.  THAT will be the square footage of your octagon.\n\nWhat... did you think I was going to do it ALL for you?
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enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
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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.