Application of Differentiation - Maximum/Minimum & Rate of Change

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.
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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3 Responses to Application of Differentiation - Maximum/Minimum & Rate of Change

  1. 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. 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. 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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