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Application of Differentiation - Maximum/Minimum & Rate of Change


Qn2
A closed box with a square base of length x and height h, is to have a volume, F, of 150m^3. The material for the top and bottom of the box costs $2 per square metre, and the material for the sides of the box costs $1.50 per square metre. Fnd the value of x and h, correct to 3 decimal places, if the total cost of the materials, C is to be a minimum.
Qn3
A viscous liquid is poured onto a flat surface. It forms a circular patch which grows at a steady rate of 6cm^2/s. Find,
a) the radius r, in pie, of the patch 24 seconds after pouring has commenced.
b) the rate of increase of the radius at this instant, correct to 2 decimal places.

Done reading the qns? Do and see if your answers are correct!

:)

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Ai Ling Ong

Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)

Filed Under: A-Maths Tuition Tagged With: differentiation, maximum minimum, rate of change

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Comments

  1. lijun(: says

    October 2, 2007 at 12:18 am

    i came across this qu that day, and i really want to clarify if the method i used is correct or not.

    int(tan^2x - x^3)

    i changed it to int(sec^2x - 1 - x^3) instead, cos i don't think we can integrate tan.

    is that feasible?

  2. hmm says

    April 14, 2008 at 12:54 am

    For Qn2,

    Total Cost, C =4x^2 + 6xh,

    Shouldn't it be C =4x^2 + 4xh instead?

    Maybe the question could also add in the total cost for the box

  3. alwaysLovely says

    April 14, 2008 at 9:49 am

    The material for the top and bottom of the box costs $2 per square metre, and the material for the sides of the box costs $1.50 per square metre

    ==> Total cost for top & bottom of box = 2*2x^2 =4x^2
    ==> Total cost for sides of box = 1.5*4xh =6xh
    ==> Total cost = 4x^2 + 6xh
    :)

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