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Amanda [17]
2 years ago
5

Find positive numbers x and y satisfying the equation xyequals12 such that the sum 4xplusy is as small as possible. Let S be the

given sum. What is the objective function in terms of one​ number, x?
Mathematics
1 answer:
WITCHER [35]2 years ago
4 0

Answer:

The objective function in terms of one​ number, x is

S(x) = 4x + (12/x)

The values of x and y that minimum the sum are √3 and 4√3 respectively.

Step-by-step explanation:

Two positive numbers, x and y

x × y = 12

xy = 12

S(x,y) = 4x + y

We plan to minimize the sum subject to the constraint (xy = 12)

We can make y the subject of formula in the constraint equation

y = (12/x)

Substituting into the objective function,

S(x,y) = 4x + y

S(x) = 4x + (12/x)

We can then find the minimum.

At minimum point, (dS/dx) = 0 and (d²S/dx²) > 0

(dS/dx) = 4 - (12/x²) = 0

4 - (12/x²) = 0

(12/x²) = 4

4x² = 12

x = √3

y = 12/√3 = 4√3

To just check if this point is truly a minimum

(d²S/dx²) = (24/x³) = (8/√3) > 0 (minimum point)

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Answer:

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Step-by-step explanation:

27m^3+125n^3=\left(3m\right)^3+\left(5n\right)^3

Apply sum of cubes formula: x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)

\left(3m\right)^3+\left(5n\right)^3=\left(3m+5n\right)\left(3^2m^2-3m\times 5n+5^2n^2\right)

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Step-by-step explanation:

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