1. A rectangular box is to be constructed from 2 different materials. The box will have a square base and open top. The material for the bottom costs $4.25/m2. The material for the sides costs $2.5/m2. Find the dimensions of the box with the largest volume if the budget is $500 for the material.
2. The concentration, C, of a drug injected into the bloodstream t hours after injection can be modeled by C(t) = (t/4) + 2t^-2. Determine when the concentration of the drug is increasing and when it is decreasing.
Let the base of the box by x by x m, and let the height be y m.
visualize the box flattened out, the base is a square, with an x by y rectangle attached to each of its sides.
Your "cost" equation is
500 = 4.25x^ + 4*2.5xy
solve this for y
Your main equation, the part that is to be maximized, is
Volume = x^2*y
plug in the y from the other equation, simplify, then differentiate
set the derivative equal to zero and solve.
This is a standard and easy optimization problem and you should not have any difficulty with it.
for the second one, "something" is increasing when its first derivative is positive, and "decreasing" when its first derivative is negative.
you should be able to take it from there.
1. To find the dimensions of the box with the largest volume, you need to set up an equation that represents the budget constraint and the volume equation.
Let's assume the base of the box has dimensions of x by x meters, and the height of the box is y meters.
The cost of the material for the bottom (base) is given as $4.25/m^2, and the cost of the material for the sides is $2.5/m^2.
The cost equation is then given by:
500 = 4.25x^2 + 4 * 2.5xy
Solve this equation for y:
500 - 4.25x^2 = 10xy
y = (500 - 4.25x^2) / (10x)
Now, we need to express the volume equation using this value of y:
Volume = x^2 * y
Substituting the value of y, we get:
Volume = x^2 * ((500 - 4.25x^2) / (10x))
Simplifying the expression, we get:
Volume = (50x - 0.425x^3) / 10
To find the dimensions of the box with the largest volume, we need to differentiate the volume equation with respect to x, set it equal to zero, and solve for x.
d/dx (Volume) = 0
d/dx ( (50x - 0.425x^3) / 10 ) = 0
Differentiating with respect to x:
(50 - 1.275x^2) / 10 = 0
Solving for x:
50 - 1.275x^2 = 0
1.275x^2 = 50
x^2 = 50 / 1.275
x^2 = 39.2157
x ≈ 6.26
Since x represents the length of the side of the base, we can round it to the nearest whole number and assume x = 6.
Now, substitute the value of x into the equation for y to find the corresponding value:
y = (500 - 4.25 * 6^2) / (10 * 6)
y ≈ 8.41
Therefore, the dimensions of the box with the largest volume within the budget of $500 are approximately 6m by 6m by 8.41m.
2. To determine when the concentration of the drug is increasing or decreasing, we need to analyze the first derivative of the concentration equation.
Given the concentration equation: C(t) = (t/4) + 2t^-2
We need to find the first derivative, which represents the rate of change of the concentration with respect to time:
dC/dt = d/dt (t/4) + d/dt (2t^-2)
Differentiating each term individually:
dC/dt = 1/4 + (-4t^-3)
Simplifying:
dC/dt = 1/4 - 4/t^3
The concentration is increasing when the first derivative is positive, and it is decreasing when the first derivative is negative.
Setting the first derivative equal to zero:
1/4 - 4/t^3 = 0
Multiply both sides by 4t^3:
1 - 16 = 0
-15 = 0
However, this equation has no real solutions. Therefore, the concentration is never at a maximum or minimum, which means it is either always increasing or always decreasing.
To determine which, we can analyze the first derivative for different values of t. If the first derivative is positive, the concentration is increasing, and if it's negative, the concentration is decreasing.
For example, if we plug in t = 1:
dC/dt = 1/4 - 4/1^3 = 1/4 - 4 = -15/4
Since the first derivative is negative (-15/4 < 0), we can conclude that the concentration is decreasing for all values of t.