Given that xy =3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + 3y/5. Please show steps. Thank you!​

Question

Given that xy =3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + 3y/5. Please show steps. Thank you!​

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bonexptip 4 years 2021-07-21T12:28:57+00:00 1 Answers 11 views 0

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  1. Sapo1
    0
    2021-07-21T12:30:01+00:00
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    First of all, x and y cannot be zero because of the condition xy = 3/2. Hence x and y can only be positive.

    Now let F(x,y)=10x+\frac{3y}{5}

    Since y=\frac{3}{2x}, ==> F(x,y) becomes, after substitution,

    F(x)=10x+\frac{9}{10x}

    ==> F'(x) = 10-\frac{9}{10x^{2} }

    Setting this to 0 to find the critical point, we get 10=\frac{9}{10x^{2} }

    or x^{2} = \frac{9}{100}

    ==> x = 3/10. (Chose the positive because of the hypothesis.)

    ==> y = 5.

    Now the 2nd derivative is F” = \frac{9}{5x^{3} } > 0

    hen x = 3/10.

    Hence an (absolute) minimum exists at (3/10,5), with value

    F(\frac{3}{10} , 5) = 10 *\frac{3}{10} +\frac{3*5}{5} =6

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