1. ## Quadratic Equation

Ok so i have been given an asignment to work out x in this question...

I have changed the values from their original values so as to not have my asignment done for me...

how much i go about finding the value of x?

Thank you.

2. Originally Posted by Fitzgerald
Ok so i have been given an asignment to work out x in this question...

I have changed the values from their original values so as to not have my asignment done for me...

how much i go about finding the value of x?

Thank you.
You don't- unless there is more information given.

Volume is "width times height times length" and here, the width is 3- 2x, the height is x, and the length is 5- 2x. Thus, the volume is given by V= (3- 2x)(x)(5- 2x) which is a cubic equation, not a quadratic. IF you are given a specific volume, then you could solve that cubic equation for x.

It is also true that the area of a rectangle is "width times length" and the surface of this box consists of 5 rectangles, two with length= 5-2x and width x, two with length 3- 2x and height x, and one with length 5-2x and width 3-2x. The total surface area of the box is A= 2(5-2x)(x)+ 2(3-2x)(x)+ (5-2x)(3-2x) which, if you multiply it out is a quadratic equation. If you have a specific number for the surface area, then you could solve for x.

Now, what does the problem really ask you to do?

3. Well if that is the case then here is the original question that followed this information.

Calculate the value of X and hence the dimensions of the crate to give the maximum volume of the crate using both algebraic as well as graphical methods. You may use the fact that the gradient at maximum turning point is zero.
That is, if y=ax^n then