A rectangular field with area 5000^2 is enclosed by 300 of fencing. Find the dimensions of the field.?

perimeter = 2w + 2L = 300

w + L = 150

w = 150 - L

if we let the length be x, then width = (300 - 2x) / 2 = 150 - x

(draw yourself a sketch to convince yourself of this)

Area = length * width = x ( 150 - x) = 5000

150x - x^2 = 5000

x^2 - 150x + 5000 = 0

(x - 50)(x - 100) = 0

so x = 50 or x = 100

the field is 50 by 100

check: 2(50) + 2(100) = 300 feet of fence -- yep

area = 50* 100 = 5000 check

once you do optimization complications, the first ingredient is to ask your self, what's being optimized? in this situation, you're requested for the most important area. ok, so as which ability you want a formula for area. What structure are you operating with? in this situation, a oblong field. So say it has length L and width W, then its area is A = L*W next step is to locate the constraint equation so that you'll be able to remedy for L in words of W (or vice versa). The constraint right that is that your value is constrained to $1000. call value "C". C = 1000 yet what's a formula for C? properly it truly is the position all the stuff interior the middle is presented in. Fence cloth expenditures $2 in holding with foot for both ends and $4 in holding with foot for the parallel area. So if each and each end is W feet lengthy and the parallel area is L feet huge, the cost is: C = 2*W + 2*W + 4*L (2 income holding with W feet for both aspects and four income holding with L feet for the single area) Now on the grounds that C = 1000, we get 1000 = 2*W + 2*W + 4*L which simplifies to 1000 = 4*W + 4*L now you remedy for W in words of L: 1000 - 4*L = 4*W 250 - L = W and plug decrease back into unique equation for W: A = L*W A = L*(250 - L) A = 250*L - L^2 (simplifying) Now that you've the equation in words of one variable in worry-free words, do your regular optimization stuff. Take 1st by-product of A, A' = 250 - 2*L, set it equivalent to 0 and remedy for L to get the serious values. (One subsequently.) Do both the first or 2d by-product attempt to be positive that's a max, and that'll be the L-length on your answer. you're requested to provide both dimensions, so once you get the L which maximizes area, bypass decrease back to 250 - L = W and plug interior the maximal L you got here upon to get the W length besides.