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3 years ago

A peach orchard owner wants to maximize the amount of peaches produced by her orchard.

Mathematics

1 answer:

qaws [65]3 years ago

6 0

Answer:

a) Y(x) = {900, x≤30; 900-40(x-30), x>30}

b) T(x) = {900x, x≤30; 2100x-40x², x>30}

c) dT/dx = {900, x≤30; 2100-80x, x>30}

Step-by-step explanation:

a) The problem statement gives the function for x ≤ 30, and gives an example of evaluating the function for x = 35. So, replacing 35 in the example with x gives the function definition for x > 30.

$\displaystyle Y(x)=\left\{\begin{array}{lcl}900&\text{for}&x\le 30\\900-40(x-30)&\text{for}&x>30\end{array}\right.$

b) The yield per acre is the product of the number of trees and the yield per tree:

T(x) = x·Y(x)

$\displaystyle T(x)=\left \{\begin{array}{lcl}900x&\text{for}&x\le 30\\2100x-40x^2& \text{for}&x>30\end{array}\right.$

c) The derivative is ...

$\displaystyle\frac{dT}{dx}=\left \{\begin{array}{lcl}900&\text{for}&x\le 30\\2100-80x& \text{for}&x>30\end{array}\right.$

_____

The attached graph shows the yield per acre (purple, overlaid by red for x<30), the total yield (black), and the derivative of the total yield (red). You will note the discontinuity in the derivative at x=30, where adding one more tree per acre suddenly makes the rate of change of yield be negative.

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2 years ago

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Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

$f_x = ycos(xy)$

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we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

$\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)$

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

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if we derivate f over z, we obtain

$f_z(x,y,z) = -ye^{yz} + d/dz K(z)$

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B686x%5E4%20y%5E7%7D" id="TexFormula1" title="\sqrt[3]{686x^4 y^7}" alt="\sqrt

Sergio [31]

This answer is there in picture