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Application of Differentiation – Maximum/Minimum & Rate of Change
3 Comments | Posted by alwaysLovely in A-Math
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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3 Comments for Application of Differentiation – Maximum/Minimum & Rate of Change
lijun(: | October 2, 2007 at 12:18 am
hmm | 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
Author comment by alwaysLovely | 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
:)
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?
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