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Our Maths Genius Should Help O / Please, Maths Gurus In The House, Help With Solution / My Physics/further Maths Gurus. Check This Out (2) (3) (4)
Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 12:27pm On Jan 13, 2019 
Good afternoon my able maths gurus. I love this forum so much in that i have learn alot from you guys and also appreciate ur efforts and contributions to this great forum. So, my kid brough this to me and i was like., is this girl trying to crack my brain? A sheet metal is used for making a closed rectangular box with square ends. If the total area of the metal is 216cm², find the dimension that will give the greatest volume. 2. From the result of the soil test carried out on the school farm ear marked for cowpea production, it was discovered that 37kg of phosphorus per heitare will be needed to ensure good yield. If single superphosphate fertilizer is to use as the source of phosphorus. Calculate the amount or weight of single superphosphate fertilizer that will be required to supply the amount of phosphorus required. (single superphosphate fertilizer contains 18% phosphorus, a bag is 100kg). Thanks 
Re: Our Maths Gurus Please Come In And Solve This For Me by sparog(m): 2:35pm On Jan 13, 2019 
Kelly2713: This one na maths or soil chemistry? 1 Like 
Re: Our Maths Gurus Please Come In And Solve This For Me by FromZeroToHero(m): 3:02pm On Jan 13, 2019 
how much will you pay? 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 3:02pm On Jan 13, 2019 
sparog:No. 1 is maths 2. Is soil 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 3:03pm On Jan 13, 2019 
Re: Our Maths Gurus Please Come In And Solve This For Me by kelvin1191(m): 3:57pm On Jan 13, 2019 
No 2 is 2.06 bag. since a bag is 100kg and contains 18% P. 18/100 *100 gives 18KG OF P IN A BAG. IF 37KG IS REQUIRED 37/18=2.055 bag or 2bag 5.5kg or 205.5kg 
Re: Our Maths Gurus Please Come In And Solve This For Me by Nobody: 4:04pm On Jan 13, 2019 
Question 1 involves calculus: The box consists of 4 rectangles and 2 squares.From this info,we can work out the formula for the surface area of the box: The surface area of a rectangle is:xh.since there are 4 rectangles,the total surface area would be 4xh. For the 2 squares,the total area is 2x^2. Therefore the total surface area for the box is: 4xh+2x^2=216.......[1] volume of box is x^2h......[2] Make h the subject of formula in equation 1 and substitute in equation 2. h=2162x^2/4x.......[3] Therefore V=x^2(2162x^2/4x)=54x(x^3/2)....(4)on simplification. According to rule of calculus,The volume is maximum when dV/dx=0 on differentiating equation 4, We get dV/dx=54(3x^2/2)=0.....(5) Solve for x in equation 5 54=3x^2/2 3x^2=108 X^2=36 x=6 Now we have gotten x,substitute its value in equation 3 to get h:h=2162(6)^2/4(6) h=6 So the dimension that can produce a box of maximum volume is(6,6) I.e Two squares with a length of 6 cm each and 4 rectangles with a length of 6cm and a breadth of 6cm each are what is required for a box of maximum volume. 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 5:49pm On Jan 13, 2019 
Kolombz: 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kylekent59: 6:01pm On Jan 13, 2019 
Kolombz: He made a mistake at the Bolded h= 216/4x  2x/4. Better inform the guy 
Re: Our Maths Gurus Please Come In And Solve This For Me by Nobody: 6:11pm On Jan 13, 2019 
Kylekent59:There is nothing wrong with the equation.What I meant is actually (2162x^2)/4x.you went ahead to divide each component of the numerator(216 and 2x^2) with the denominator.But your answer is wrong. Since you are dividing 2x^2 by 4x,the result is supposed to be 2x/4 since x cancels x not 2x^2/4 like you wrote in your correction.Thanks for your observation btw. 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kylekent59: 6:29pm On Jan 13, 2019 
cool 1 Like

Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 6:51pm On Jan 13, 2019 
Kylekent59:Thanks bro 
Re: Our Maths Gurus Please Come In And Solve This For Me by fr3do(m): 7:42pm On Jan 13, 2019 
.. 1 Like

Re: Our Maths Gurus Please Come In And Solve This For Me by ikbnice(m): 7:53pm On Jan 13, 2019 
What class is your kid if I may ask 1 Like 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 8:34pm On Jan 13, 2019 
ikbnice: Ss3 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 8:36pm On Jan 13, 2019 
ikbnice:Ss3 bro 
Re: Our Maths Gurus Please Come In And Solve This For Me by pyrex23(m): 8:57pm On Jan 13, 2019 
This is a 100l question.... Well.. Since it has been solved.. 
Re: Our Maths Gurus Please Come In And Solve This For Me by Nobody: 9:24pm On Jan 13, 2019 
sparog: 
Re: Our Maths Gurus Please Come In And Solve This For Me by StevenOba: 9:27pm On Jan 13, 2019 
Oh this is easy na. Last last the answer is the answer. Thank you. 
Re: Our Maths Gurus Please Come In And Solve This For Me by 4Mea: 11:45pm On Jan 13, 2019 
That one na maths 
Re: Our Maths Gurus Please Come In And Solve This For Me by rtdCivilservant: 6:58pm On Jan 14, 2019 
Kolombz:oga u failed it. the dimension that would give u max volume=216×1×1. QED 
Re: Our Maths Gurus Please Come In And Solve This For Me by rtdCivilservant: 7:00pm On Jan 14, 2019 
Kelly2713:the answer to ur number 1 question =216×1×1. simple. QED any other answer is totally wrong and unacceptable 
Re: Our Maths Gurus Please Come In And Solve This For Me by Martinez39(m): 11:58am On Feb 14, 2019 
Kelly2713:1) The total area of the metal sheet is equal to the total area of the box. Given that the box has square ends and rectangular sides the the surface area SA of the closed box is SA = 4yx + 2x² = 216 , this reduces to 2yx + x² = 108 . 2) The volume V of the box is given by V = V(x,y) = x²y . Solving for y in the surface area gives y = (108  x²) ÷ 2x . Substituting this into the volume equation yields V = V(x) = 54x  (½)x³ . This is the equation to be maximised and by logic, the domain of x is the open interval 0 < x < (216)^½. 3) To find the value of x that yields the maximum volume, you need to find the critical number(s) of the volume function. The critical number for this equation will only occur at x  values for which dV/dx = V'(x) = 0 . Given that dV/dx = V'(x) = 54  2(¾)x² , the x  values for which dV/dx = V'(x) = 0 are ±6 , we pick 6 because it lies in the feasible domain of x . So the value of x that maximises V = V(x) is 6cm. 4) to find the corresponding value of y that maximises V = V(x) = V(x,y) , put the value of x that maximises V= V(x) into the surface area equation and solve for y . Doing this yields y = 6cm . 5) So if the square ends of both boxes are 6cm × 6cm and the four rectangular sides are of the dimensions 6cm × 6cm , you will get the maximum volume. 
Re: Our Maths Gurus Please Come In And Solve This For Me by SirWarlock: 12:46pm On Feb 14, 2019 
Maths Too bad I got into life science 
Re: Our Maths Gurus Please Come In And Solve This For Me by Kelly2713(m): 6:22pm On Feb 14, 2019 
Martinez39:Thanks 
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