Question 1163322
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Let the width be x.<br>
Since the corral is divided in half, there are three fences of length x; that's a total fence length of 3x.<br>
The remaining length of fence, 120-3x, is the two other sides of the corral, so the length of the corral is (120-3x)/2, or 60-1.5x.<br>
The area of the corral is then<br>
{{{x(60-1.5x) = -1.5x^2+60x}}}<br>
This is a quadratic expression of the form ax^2+bx+c; its value is maximized when x is equal to -b/2a.<br>
In this quadratic expression, {{{-b/(2a) = -60/-3 = 20}}}<br>
So the width that maximizes the volume is x=20 (which satisfies the condition that it must be at least 6).  And then the length is (120-3x)/2 = 30.<br>
ANSWER: length 30 and width 20 maximizes the area.<br>
Note that the total lengths of fencing in the two directions are equal.  This is always the case, no matter how many adjacent corrals the whole corral is divided into.<br>
For example, if you have 600 feet of fencing and you are dividing the long corral into 5 adjacent sections (using 6 sections of fence width-wise), then the maximum area is with 300 feet of fencing in each direction -- which means a length of 300/2 = 150 feet and a width of 300/6 = 50 feet.<br>
So to an experienced problem solver, the solution to your problem would go like this:<br>
Total fence length: 120 feet
Length of fencing in each direction: 120/2 = 60 feet
Length of corral for maximum area: 60/2 = 30 feet
Width of corral for maximum area: 60/3 = 20 feet<br>