A rancher has 1000m of fencing to enclose two rectangular corrals. The corrals h

Question

A rancher has 1000m of fencing to enclose two rectangular corrals. The corrals have the same dimensions and one side in common. What dimensions will maximize the enclose area? ____________ |_____|_____| y . x THIS is all i've solved for so far my two formulas are 1) 1000=4x+3y 2) A=2LW two given that there are 2 rectangles. isolate the y in the first equation (1000-4x)/3 = y plug this equation into the 2nd function A=2x(1000-4x)/3 find derivative A = (2000x-8x^2)/3 A'= (2000-16x)/3 factor out 16 A'=16(125-x)/3 x=125m

Explanation / Answer

Your last step will be that you set A' to zero, and you get x = 125 Thus, x = 125 will be your critical value. Whenever you are finding the maximum or minimum value (maximum area in this case), you will have to remember that such values will only exists at critical value or at boundary values. The boundary values are the maximum possible and minimum possible values of x. For this question, they are not really applicable, since y will be zero if x is at its maximum, and x will be zero if y is at its maximum. Thus, the maximum area can only occurs at a critical value. This is where the first and second derivative test comes in. If you choose to use to use the first derivative test, you will need to show that the first derivative is positive for all x-values less than 125, and the first derivative is negative for all x-values more than 125. The second derivative test is much easier to use, so most people will use it and ignore the first derivative test. If the second derivative is positive for x = 125, that means x = 125 will result in a minimum area. If the second derivative test is negative for x = 125, that means x = 125 will result in a maximum area. Note that for cases such as this, you can be assured that x = 125 will result in the maximum area even before doing any calculations, as it is the only critical value, and you have no other choices. However, you are expected to still do the workings to show that it is the value which will result in a maximum area. A' = (16/3)(125 - x) A'' = -16/3 < 0 Since A'' is negative for x = 125 (Well, it is negative for all real x-values), x = 125 is a local maximum, and thus will result in the maximum area.

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