1

Regression equation

x <- c(5.6, 6.3, 7, 7.7, 8.4)
y <- c(8.8, 12.4, 14.8, 18.2, 20.8)
lm_fit <- lm(y ~ x)
summary(lm_fit)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##     1     2     3     4     5 
## -0.24  0.38 -0.20  0.22 -0.16 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -14.8000     1.0365  -14.28 0.000744 ***
## x             4.2571     0.1466   29.04 8.97e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.3246 on 3 degrees of freedom
## Multiple R-squared:  0.9965, Adjusted R-squared:  0.9953 
## F-statistic: 843.1 on 1 and 3 DF,  p-value: 8.971e-05

\[ y = -14.8 + 4.26 x \]

2

Critical points (minima, maxima, saddle points)

\[ f(x, y) = 24x - 6xy^2 - 8y^3 \]

Find partial derivatives:

\[ f_x(x, y) = 24 - 6y^2 \] \[ f_y(x, y) = -12xy - 24y^2 \] Set both equal to zero and solve.

\[ y^2 = 4 ; y = \pm 2 \] \[ x = -4 \] Plug into original equation to get:

(-4, 2, -64) and (-4, -2, 64)

3

Step One

Take the two formulas for units sold, multiply them each by dollars, and add them.

\[ revenue(x, y) = x(81 - 21x + 17y) + y(40 + 11x - 23y) \] \[ = 81x - 21x^2 + 17xy + 40y + 11xy - 23y^2 \] \[ = -21x^2 - 23y^2 + 28xy + 81x + 40y \]

Step Two

\[ revenue(2.3, 4.1) = 116.62 \]

4

Minimize costs by finding the partial derivatives of the cost function, then setting them equal to each other, and substituting back into the constraint formula.

\[ C(x, y) = \frac{1}{6} x^2 + \frac{1}{6} y^2 + 7x + 25y + 700 \] \[ C_x(x,y) = \frac{1}{3} x + 7 \] \[ C_y(x,y) = \frac{1}{3} y + 25 \] \[ \frac{1}{3} x + 7 = \frac{1}{3} y + 25 \] \[ x = y + 54 \]

Constraint formula:

\[ x + y = 96 \] \[ y + 54 + y = 96 \] \[ y = 21 \] \[ x = 75 \]

5

Double integral

\[ \int_2^4 \int_2^4 (e^{8x + 3y}) dy dx \] \[ = \int_2^4 ([\frac{1}{3}e^{8x + 3y}] \rvert_2^4) dx \] \[ = \int_2^4 (\frac{1}{3}e^{8x + 12} - \frac{1}{3}e^{8x + 6}) dx \] \[ = \int_2^4 (\frac{1}{3}e^{8x}(e^{12} - e^{6})) dx \] \[ = (\frac{1}{24}e^{8x}(e^{12} - e^{6})) \rvert_2^4 \] \[ = (\frac{1}{24}e^{32}(e^{12} - e^{6}) - \frac{1}{24}e^{16}(e^{12} - e^{6})) \] = 534,155,900,000,000,000