# Math Help - Application of Differentiation..maximum and minimum values

1. ## Application of Differentiation..maximum and minimum values

Note
Hello,..
I can't find a solution for this question can anyone help plz..
here is the question:

A box is constructed out of a 3x4 m2 (metre square) rectangular piece of cardboard by cutting away at each edge a square of side length x and folding up each edge.

(a) Find an expression of the box's volume in terms of x.
(b) What are the possile values of x.
(c) Determine the value of x which maximizes the volume of the box.
Thanking you all in advance for helping...

2. Originally Posted by NiCeBoY
Note
Hello,..
I can't find a solution for this question can anyone help plz..
here is the question:

Thanking you all in advance for helping...
Have you tried drawing a net for the box ....? Clearly the height of the box is x. What are the dimensions of the base .....?

3. You need to find the volume and then take the derivative of the volume and then equal the derivative to 0 which will give you the critical points.

$A=(3-2x)(4-2x)$

$V=x(3-2x)(4-2x)$

$\frac{dv}{dx} = 12x^2 - 28x + 12$

$12x^2 -28x + 12 = 0$

4. on solving:
$12x^2 -28x + 12 = 0$

i get x=0.57 and x=1.78

is this the possible values of x?

and if x=1.78,is it when the voluma of the box is maximum plz.?

thx

5. x can't be greater than 1.5 ... can you reason why?

6. yes because if one side of the card board is 3m and (2x1.5) should be remove so there will be nothing left..

so how i should proceeed plz.?

7. you have another solution ... how can you check that it yields a maximum?

8. when equating dv/dx with o if the value of x is negative its maximum and its positive its a minimum.. i think..

9. Try using the second derivative test to see if it a max

10. when i equate the second derivative of :
$12x^2 -28x + 12$

which is

$24x - 28$

to 0 the value of x is 1.17 which is positive. so i think it should be a minimum..

11. Try again 1.17 is a point of inflection. You on the right track but instead of equaling the second derivative to 0 just plug in your critical point from the first derivative into the second derivative. If it gives you a positive answer it a minimum and if produces a negative answer it is a maximum.

12. that is i use the value of x=0.57 and x=1.78 and see if its positive or negative? am i right?

13. You can't use 1.78 for the reason you stated earlier. Use .57

$-14.32 = 24(.57) - 28$

The 2nd derivative gives a negative answer so it is a maximum.

Hope this helps.

14. Then how do i answer (b) What are the possile values of x.

i think the answer is : 0< X < 1.5

Am i rite plz?

15. Originally Posted by NiCeBoY
Then how do i answer (b) What are the possile values of x.

i think the answer is : 0< X < 1.5

Am i rite plz?
yup

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