We are given the equations:
<span>5 x + 11 y = 49 -->
eqtn 1</span>
<span>u = 3 x^2 + 6 y^2 -->
eqtn 2</span>
Rewrite eqtn 1 explicit to y:
11 y = 49 – 5 x
<span>y = (49 – 5x) / 11 -->
eqtn 3</span>
Substitute eqtn 3 to eqtn 2:
u = 3 x^2 + 6 [(49 – 5x) / 11]^2
u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]
u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121
u = 4.24 x^2 – 24.3 x + 119.06
Derive then set du/dx = 0 to get the maxima:
du/dx = 8.48 x – 24.3 = 0
solving for x:
8.48 x = 24.3
x = 2.87
so y is:
y = (49 – 5x) / 11 = (49 – 5*2.87) / 11
y = 3.15
Answer:
<span>George will choose some of each commodity but more y than
x.</span>
16(2+5) since when you use distributive property, 16*2= 32 and 16*5= 80
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
Answer:
13 cars per hour
Step-by-step explanation:
She can vacuum 104 cars in 8 hours, so divide 104 by 8, 104/8, to get 13. So she can vacuum 13 cars in 1 hour.
Answer:
(11/2, -1)
Step-by-step explanation:
