Answer:
The two numbers are:
x = 500
y = 125
Step-by-step explanation:
We want to find two numbers x and y, such that:
x + 4*y = 1000
f(x, y) = x*y is maximum.
From the first equation, we can isolate one of the variables to get
x = 1000 - 4y
now we can replace it in f(x, y):
x*y = (1000 - 4*y)*y = 1000*y - 4*y^2
So now we want to maximize the function:
f(y) =- 4*y^2 + 1000*y
where y must be an integer.
Notice that this is a quadratic equation with a negative leading coefficient (so the arms of the graph will open downwards), thus, the maximum will be at the vertex.
Remember that for a general quadratic equation:
y = a*x^2 + bx + c
the x-value of the vertex is:
x = -b/(2*a)
so, in the case of:
f(y) =- 4*y^2 + 1000*y
the y-value of the vertex will be:
y = -1000/(2*-4) = 1000/8 = 125
So we found the value of y.
now we can use the equation:
x = 1000 - 4*y
x = 1000 - 4*125 = 1000 - 500 = 500
x = 500
Then the two numbers are:
x =500
y = 125
![\sqrt[4]{\frac{6}{5}} = \sqrt[4]{b^4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B6%7D%7B5%7D%7D%20%3D%20%5Csqrt%5B4%5D%7Bb%5E4%7D)
![\sqrt[4]{\frac{6}{5}} = b](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B6%7D%7B5%7D%7D%20%3D%20b)
to predict how many bacteria will be present after 10 hours.![b = \sqrt[4]{\frac{6}{5}}](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B6%7D%7B5%7D%7D)
![A = 3000\sqrt[4]{\frac{6}{5}}^{10} = 4732.3228 = 4732](https://tex.z-dn.net/?f=A%20%3D%203000%5Csqrt%5B4%5D%7B%5Cfrac%7B6%7D%7B5%7D%7D%5E%7B10%7D%20%3D%204732.3228%20%3D%204732%20)