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6x^3 + 48x^2 - 210x + 7 18x^2 + 96x - 210
18x^2 + 96x - 210 = 0 6(3x^2 + 16x - 35) = 0 3x^2 + 16x - 35 = 0 3x^2 + 21x -5x -35 3x(x+7) -5(x+7) (x+7)(3x-5) = 0
zero at -7 3x -5 = 0 3x = 5 x = 5/3
zero at 5/3
so -7 and 5/3 are our critical values
6(x+7)(3x-5)
-10 6(-10+7)(3(-10)-5) 6(-3)(-35) 630
-7 ( zero )
0 6(0+7)(3(0)-5) 6(7)(-5) -210
5/3 ( zero )
10 6(10+7)(3(10)-5) 6(17)(25) 2550
max at -7
min at 5/3
6*(-7)^3 + 48*(-7)^2 - 210*(-7) + 7 1771
6*(5/3)^3 + 48*(5/3)^2 - 210*(5/3) + 7 -1637/9
so, local max at -7, 1771 local min at 5/3, -1637/9
193 + (-119t^2)/(t^2+43) 193 -119 * ((t^2)/(t^2+43))
quotient rule
(f’g - fg’) / g^2 f = t^2 g = t^2+43 f’ = 2t g’ = 2t ((2t)(t^2+43) - (t^2)(2t))/(t^2+43)^2 (2t^3 + 86t - 2t^3)/(t^2+43)^2 (86t)/(t^2+43)
-119 * 86 = -10234
-10234t / (t^2+43)^2 denominator always positive cause exponents numerator always negative cause t is never gonna be negative decreasing 0, inf increasing never
(-119t^2)/(t^2+43)
degree is equal, so as t approaches inf, it approaches ratio of leading coefficients -119/1 -119
193 -119 = 74
so temp of room is 74
-x^2-2x+6 -2x-2 = 0 -2 = 2x -1 = x
critical value at -1, which is in interval. also evaluate at -6 and 5 cause its absolute extrema
-6 -(-6)^2 - 2(-6) + 6 -18
-1 -(-1)^2 - 2(-1) + 6 7
5 -(5)^2 - 2(5) + 6 -29
absolute minimum at 5, -29 absolute maximum at -1, 7
-3x^3 + 18x^2 + 10x - 3 -9x^2 + 36x + 10 -18x + 36
-18x + 36 = 0 36 = 18x 2 = x
-18(0) + 36 = 36 -18(4) + 36 = -36
concave up from -inf, 2 concave down from 2, inf
we already know point of inflection at 2 from last question -3(2)^3+18(2)^2+10(2)-3 = 65
(2,65)
revenue = demandx revenue = 14x cost = .2x^2 + 5.6x + 5 profit = revenue - cost profit = 14x - .2x^2 - 5.6x - 5 profit = -.2x^2 + 8.4x - 5 profit’ = -.4x + 8.4 -.4x + 8.4 = 0 8.4 = .4x 21 = x
but is it a max?
-.4(20) + 8.4 = .4 -.4(21) + 8.4 = 0 -.4(22) + 8.4 = -.4
yep max at 21
-.2x^2 + 8.4x - 5 -.2(21)^2 + 8.4(21) - 5 83.2
area = xy 1269 = xy y = 1269/x area = x*(1269/x)
extfence = (2)(14.85)(x) + (2)(14.85)(1269/x) extfence = 29.7x + 29.7(1269/x) extfence = 29.7x + 37689.3/x intfence = 11(2x) intfence = 22x cost = extfence + intfence cost = 29.7x + 37689.3/x + 22x cost = 51.7x + 37689.3/x cost’ = 51.7 - 37689.3x^-2 37689.3/x^2 = 51.7 37689.3 = 51.7x^2 729 = x^2 27 = x
so, x is 27 1269/27 = 47 y is 47
the dimensions that minimize the cost are x=27 and y=47
The maximum value of profit is 83.2$
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