Lesson Calculus optimization problems for shapes in 2D plane
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<H2>Calculus optimization problems for shapes in 2D plane</H2> <H3>Problem 1</H3>A rectangular garden is to be surrounded by a low fence. There is only 100 feet of fencing available for all four sides. What is the largest possible area of the garden that can be fenced? <B>Solution</B> <pre> If x is the length of the garden, then its width is {{{100/2-x}}} feet, and the area is A(x) = {{{x*(50-x)}}} = {{{-x^2 + 50x}}} They want you find the maximum of the function A(x) using Calculus. For it, differentiate A(x) over x and equate the derivative to zero: A'(x) = -2x + 50 = 0, which gives you x = {{{50/2}}} = 25 feet. Thus you obtain the <U>ANSWER</U>. Under given conditions, the maximum area is achieved for the square of the side length equal to one fourth of the given perimeter, i.e. 25 feet. </pre> The fact that under given conditions, the maximum area is achieved for the square, is VERY WELL known and can be obtained by means of Algebra, too. See the lessons - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-complete-the-square-of-a-quadratic-function-to-find-its-minimum-maximum.lesson>HOW TO complete the square to find the minimum/maximum of a quadratic function</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Briefly-on-How-to-complete-the-square-of-a-quadratic-function-to-find-its-minimum-maximum.lesson>Briefly on finding the minimum/maximum of a quadratic function</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-complete-the-square-to-find-the-vertex-of-a-quadratic-function.lesson>HOW TO complete the square to find the vertex of a parabola</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Briefly-on-finding-the-vertex-of-a-parabola.lesson>Briefly on finding the vertex of a parabola</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/A-rectangle-with-the-given-perimeter-which-has-the-maximal-area-is-a-square.lesson>A rectangle with a given perimeter which has the maximal area is a square</A> - <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/A-farmer-planning-to-fence-a-rectangular-garden-to-enclose-the-maximal-area.lesson>A farmer planning to fence a rectangular garden to enclose the maximal area</A> in this site. <H3>Problem 2</H3>A rectangular pen is being constructed with 500 feet of fencing. One side of the pen is the wall of a barn and does not require fencing. The other three sides are fenced. What are the dimensions of the pen that will maximize the area of the pen? <B>Solution</B> <pre> Let x be the dimension of the pen (in feet) perpendicular to the wall. Then the side parallel to the wall is (500-2x) feet long, and the area of the pen is A(x) = x*(500-2x) = -2x^2 + 500x square feet. They want you find the maximum of the function A(x) using Calculus. For it, differentiate A(x) over x and equate the derivative to zero: A'(x) = -4x + 500 = 0, which gives you x = {{{500/4}}} = 125 feet. Thus you obtain the <U>ANSWER</U>. Under given conditions, the maximum area is achieved for the rectangle with the short side of 125 ft perpendicular to the wall and long side of 500 - 2*125 = 250 ft parallel to the wall. </pre> <H3>Problem 3</H3>What is the minimum fencing length needed to construct a rectangle coral of the area 1800 ft^2 adjacent to a river using the river as a natural boundary on one side? <B>Solution</B> <pre> Let x be the coral length in the direction parallel to the river and y be the coral width in the perpendicular direction. Then your task is to minimize (x+2y) under the given condition/restriction on the area xy = 1800 (1) From (1), x = {{{1800/y}}}, so we need to minimize the function f(y) = {{{1800/y + 2y}}}. The derivative f'(y) = -{{{1800/y^2}}} + 2. To find the minimum of f(y), equate its derivative to zero -{{{1800/y^2}}} + 2 = 0 {{{1800/y^2}}} = 2 {{{y^2}}} = {{{1800/2}}} = 900 y = {{{sqrt(900)}}} = 30. <U>ANSWER</U>. The minimum fencing is at y = 30 ft perpendicular to the river and x = {{{1800/y}}} = {{{1800/30}}} = 60 ft along the river. </pre> <H3>Problem 4</H3>If x and y are two positive real numbers whose product is 100, x*y = 100, then the the minimum value of x+y is achieved at x = y = 10 m. In other words, if a rectangle has the given area of 100 m^2, then the minimal perimeter is achieved for the square of the side d = {{{sqrt(100)}}}. <U>Solution</U> It is simple calculus problem. <pre> If x*y = 100, then y = {{{100/x}}}, and the sum x+y is x + {{{100/x}}}. This function of "x", f(x) = x + {{{100/x}}} achieves the minimum when its derivative is equal to zero f'(x) = 1 - {{{100/x^2}}} = 0. Then {{{x^2}}} = 100; hence x = {{{sqrt(100)}}} = 10. Thus we proved that if x*y = 100, then x+y has minimum at x = y = 10. </pre> <H3>Problem 5</H3>A rectangular area of 1,050 square feet is to be enclosed by a fence, then divided down the middle by another piece of fence. The fence down the middle costs $0.50 per running foot, and the other fence costs $1.50 per running foot. Find the minimum cost for the required fence. <B>Solution</B> <pre> From the context, the dimensions of the rectangle are not given for advance - they are unknowns and they should be found from the minimum cost condition. Let x be one dimension and y be the other dimension of the rectangle. Then the cost of the outside perimeter fence is 1.50*(2x+2y) dollars = 3*(x+y) dollars, while the cost of the fence down middle is 0.50*x dollars. Note, that I don't know now, which dimension will be the length and which be the width. When the problem will be solved, the solution will tell me it . . . So, I need minimize the function f(x,y) = 3*(x+y) + 0.5x (1) under the condition x*y = 1050. (2) From (2), express y = {{{1050/x}}} and substitute it into (1). You will get g(x) = {{{3*(x+ 1050/x) + 0.5x}}} = {{{3x}}} + {{{3150/x}}} + {{{0.5x}}}. Differentiate it over x g'(x) = {{{3}}} - {{{3150/x^2}}} + {{{0.5}}} and equate the derivative to zero. You will get 3.5 = {{{3150/x^2}}}, or 3.5x^2 = 3150 x^2 = {{{3150/3.5}}} = 900, x = {{{sqrt(900)}}} = 30. Thus the dimensions of the rectangle are 30 ft and {{{1050/30}}} = 35 ft. The fence down middle has the length of 30 ft; hence, it is parallel to the shorter side of the rectangle. </pre> <H3>Problem 6</H3>16 inches are a combined perimeter of a circle and a square. What are the dimensions of the circle and square that produce a minimum total area? <B>Solution</B> <pre> Let "x" be the square side length, and "y" be the radius of the circle. Then 4x + {{{2*pi*y}}} = 16 inches (1) (perimeter) f(x,y) = x^2 + {{{pi*y^2}}} is the function to minimize (2) (area) In other words, you should minimize (2) under the constraint (1). From (1), express x = {{{(16-2*pi*y)/4}}} = {{{4 - 0.5*pi*y}}} and substitute it into the function (2). You will find then the function to minimize in the form g(y) = {{{(4-0.5*pi*y)^2}}} + {{{pi*y^2}}} = = {{{16 - 4*pi*y + 0.25*pi^2*y^2}}} + {{{pi*y^2}}} = {{{16 - 4*pi*y + 1.25*pi^2*y^2}}}. This quadratic function of "y" has the minimum at y = " {{{-b/(2a)}}} " = {{{(4*pi)/(2*1.25*pi^2)}}} = {{{3.2/pi}}}. <U>ANSWER</U>. The values that provide the minimum of the total area are y= {{{3.2/pi}}} inches (the circle radius) and x = {{{(1/4)*(16 - 2*pi*y)}}} = {{{(1/4)*(16 - 2*pi*(3.2/pi))}}} = {{{(1/4)*(16-6.4)}}} = {{{(1/4)*9.6)}}} = 2.4 inches (the side of the square). </pre> <H3>Problem 7</H3>A rectangular cardboard poster is to have a 96 square inch rectangular section of printed material, a 2-inch border top and bottom, and a 3-inch border on each side. Find the dimensions and area of the smallest poster that meets these specifications. <B>Solution</B> <pre> They want you to minimize f(x,y) = (x-6)*(y-4) under the condition x*y = 96. (1) where x and y are the dimensions of the poster. In other words, from (1), you need to minimize f(x,y) = xy - 4x - 6y + 24 = 96 - 4x - 6y + 24 = -4x - 6y + 120. From (1), you have y = {{{96/x}}}, so the function f(x,y) takes the form g(x) = -4x - {{{6*(96/x)}}} + 120 = -4x - {{{576/x}}} + 120. (2) So, you differentiate (2), and you get g'(x) = -4 + {{{576/x^2}}}. Equate it to zero g'(x) = 0 = -4 + {{{576/x^2}}}. So, to find x, you have this equation {{{4x^2}}} = 576, {{{x^2}}} = {{{576/4}}} = 144 x = {{{sqrt(144)}}} = 12 centimeters. Then y = {{{96/12}}} = 8 cm. Thus the dimension of the poster are 12 inches width and 8 inches height. </pre> <H3>Problem 8</H3>Prove that if the sum of two non-negative real number x and y is 9, x + y = 9, then the product {{{x*y^2}}} has the local maximum value of 108. Find the values of x and y that provide this maximum. <U>Solution</U> <pre> If x+y = 9, then y = 9-x, and the function f(x,y) = {{{x*y^2}}} is f(x,y) = {{{x*y^2}}} = {{{x*(9-x)^2}}}. So, we need to find the maximum of the function of x g(x) = {{{x*(9-x)^2}}}. To find the maximum of the function g(x), take its derivative and equate it to zero. The derivative is g'(x) = {{{(9-x)^2}}} - 2x*(9-x) = {{{81 - 18x + x^2}}} - {{{(18x - 2x^2)}}} = {{{81 - 36x + 3x^2}}}. Equating it to zero, you get {{{81 - 36x + x^2}}} = 0, which is equivalent to {{{x^2 - 12x + 27}}} = 0. Left side is factorable (x-9)*(x-3) = 0. and two solution of the quadratic equation are x= 9 and x= 3. The value x= 9 provides the local minimum of the function f(x), while x= 3 provides the local maximum. At x= 3, y= 9-3 = 6, and the function g(x,y) = {{{3*6^2}}} = 108. {{{graph( 330, 330, -4, 20, -20, 120, x*(9-x)^2 )}}} Plot g(x) = x*(9-x)^2 </pre> My other additional lessons on Miscellaneous word problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/misc/I-do-not-have-enough-savings-now.lesson>I do not have enough savings now</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/In-a-jar-all-but-6-are-red-marbles.lesson>In a jar, all but 6 are red marbles</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/How-many-boys-and-how-many-girls-are-there-in-a-family.lesson>How many boys and how many girls are there in a family ?</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/What-is--the-last-digit-of-the-number-a%5En-.lesson>What is the last digit of the number a^n ?</A> - <A HREF=https://www.algebra.com/algebra/homework/word/misc/Find-the-last-three-digits-of-these-numbers-.lesson>Find the last three digits of these numbers</A> - 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