Calculus Applied Optimization Problems?
1.) Suppose you are to make a rectangular box with a square base from two different materials. The material for the top and four sides of the box costs $1/ ft^2; the material for the base costs $2/ ft^2. Find the dimensions of the box of the greatest possible volume if you are allowed to spend $144 for the material to make it. Let's define some variables :: ~ length of the base of the box = x ~ length of the height of the box = y Now, we need to assume some functions :: Volume = x^2y Cost = 2 * ft^2 of base + 1* ft^2 of sides and top Cost = 2*x^2 + 1*(4xy + x^2) Cost = 3x^2 + 4xy 144 = 3x^2 + 4xy Now, we want to maximize the Volume function ... but, we need one variable ... It would be best to substitute for "y". 144 = 3x^2 + 4xy ( 144 - 3x^2 ) / ( 4x ) = y Now, let's substitute into the Volume function. V(x,y) = x^2y V(x) = [ x^2 ][ ( 144 - 3x^2 ) / ( 4x ) ] V(x) = ( 144x - 3x^3 ) / ( 4 ) V'(x) = ( 144 - 9x^2 ) / ( 4 ) V'(x) = 0 = ( 144 - 9x^2 ) / ( 4 ) x = 4 ... and y = ( 144 - 3x^2 ) / ( 4x ) = 6 4ft base and 6ft height 2.) A straight wire 60 cm long is bent into the shape of an L. What is the shortest possible distance between the two ends of the bend wire? L ... vertical piece has a length of L and the horizontal piece has the length of 60 - L The distance is actually the "hypotenuse" of this "triangle". d^2 = L^2 + ( 60 - L )^2 = L^2 + 3600 - 120L + L^2 d^2 = 3600 - 120L + 2L^2 d = ( 3600 - 120L + 2L^2 )^(1/2) d' = (1/2)( 3600 - 120L + 2L^2 )^(-1/2)( -120 + 4L ) d' = 0 = (1/2)( 3600 - 120L + 2L^2 )^(-1/2)( -120 + 4L ) 0 = -120 + 4L 30 = L 30 and 60 cm pieces 3.) Suppose the cost incurred in operating a cruise ship for one hour is a + bv^3 dollars, where a and b are positive constants and v is the ship's speed in miles per hour. At what speed should the ship be operated between two ports to minimize the cost? Cost = a + b*v^3 C = a + b*v^3 C' = 3b*v^2 C' = 0 = v When the ship isn't moving, the cost function is minimized! However, since passengers would be upset, the boat should be moving, but it should be moving as slow as possible!!!!
Optimization (Calculus): A box has a square base (with no top). There is 200 cm^2 of material to work with.?
x = base y = height x^2 + 4xy = 200 y = [200-x^2]/4x V = volume = y*x^2 V = (200-x^2)x/4 = 50x - x^3/4 dV/dx = 50 - 3x^2/4 = 0 for maximum 200 = 3x^2 x = 8.165 cm y = 4.082 cm
Solving Optimization Problems (Calculus)?
Let width of square base be w and height h. Area of 4 sides = 4wh ----> cost of 4 sides = 2*4wh = 8wh Area of base = w^2 ----> cost of base = 4*w^2 Total cost = 4w^2 + 8wh = 1200 ----> h = (1200 - 4w^2)/8w Volume V = w^2*h = w^2 *(1200 - 4w^2)/8w = 150w - 0.5w^3 Differentiate to get dV/dw and set equal to zero. Solve the equation to find the value of w which maximises volume. you should find that w = 10.
If 1500 square centimeters of material is available to make a box with a square base and an open top, find the?
Let h be the height and x be the base. It is a box with open top and square base. Area = 4xh+x^2 = 1500 4xh=1500-x^2 h=1500/4x-x^2/4x h=375/x-x/4 Volume = x^2 h Volume = x^2[375/x-x/4] = 375x-x^3/4 Differentiate the volume with respect to x dV/dx = 375 -3x^2/4 = 0 3x^2/4 = 375 3x^2 = 1500 x^2=500 x= sqrt(500) =22.3607 cm h=375/x -x/4 =11.18 cm Largest volume = x^2 h = (22.3607)^2 (11.18) = 5590.01 cm^3 dV/dx = 375 -3x^2/4 d^2V/dx^2 = -6x/4 < 0 when x=22.36 Since the second derivative is negative when x=22.36, the volume has been maximized.
What is wrong with the Indian education system?
9th standard:PTA meetingPhysics Teacher: Your son is very smart. He is hardworking too. Dad: Thanks sir! He studies well. He is passionate about physics.Physics Teacher: Yes. He has what it takes to be an IAS Officer. Prepare him for Civil Services. He would be a great officer.Dad: That's good to hear sir. I am just a 9th standard guy. All I knew was that I loved applying physics to daily life activities. I knew I loved Maths too. But I did not know about my future. I did not know what I would be or what I wanted to be. I approached my teacher the next day.Class - the next day:Me: Sir. I like physics. What is its future.Physics Teacher: The future of physics is full of possibilities. There is infinite scope for exploring the field.Me: Then why did you want me to be an IAS Officer.Physics Teacher: Its a very stable job. You get security. You can get others to do your work for you. You will have many servants. You can earn very much. Even though the salary from government is less, you get lots of gifts (not so subtle). If you pursue physics, you can't earn much. Home:Mom: How did the meeting go. Dad: Great! His teacher told me that he has what it takes to even crack Civil Services.Mom: He's too young. Let us give him a few years before we push him towards his goals. (My mother is a professor. I am lucky. Not all are lucky to be afforded this freedom.)Conclusion:I tell my teacher I like his subject. He tells me it has good future. But money is the ultimate objective and it has less scope to earn money. He tells me to pursue another field. I feel there is something fundamentally wrong if we educate our younger generation that one field has more priority over another. All are equal. It just depends on one's passion towards a field and a drive to pursue it.As someone said: "Where would they be if Lata Mangeshkar picked up a cricket bat and Sachin Tendulkar became a singer."
Construction cost calculus question?
Call the length of the base b The volume of the box is (b ^2)h, where h is the height of the box. So we have (b^2)h = 250 from which we can get that h = 250/(b^2) The material required is a top and bottom + 4 sides. Top and bottom each have an area of b^2 Each side has an area of bh The cost for the top and bottom is 2(b^2)($2) The cost for the 4 sides is 4(bh)($1) so the total cost is 4(b^2) + 4bh since h = 250/(b^2) we can change the cost to 4(b^2) +4b(250/b^2)) = 4(b^2) + 1000/b
Optimization: minimize the cost?
front width: w depth: d height h total volume is given by wdh=200 or h=200/(wd) now the cost (in cents) is given by 8wd+15wh+8dh = 8wd + (15w+8d)h substituting with h=200/(wd) we get C(w,d) = 8wd + 200(15w+8d)/(wd) now to minimize we want to solve dC/dw = 0 and dC/dd = 0 thus 8d - 1600/w^2 = 0 8w - 3000/d^2 = 0 solving the first for d we get d = 200/w^2 substituting into the second we get 8w - 3000w^2/200 = 0 8w - 15w^2 = 0 w(8-15w) =0 w=8/15 d=5625/8 h=8/15