### Video Transcript

Determine the values of π₯ and π¦
that maximise the function π equals five π₯ plus two π¦. Write your answer as a point π₯,
π¦.

Weβve been given a region which
represents a series of inequalities. The topmost horizontal inequality
is π¦ is less than or equal to eight. The vertical inequality is π₯ is
less than or equal to seven. The region is also bound by the
π₯-axis. So π¦ must be greater than or equal
to zero. And finally, we have a diagonal
line with a π¦-intercept at eight and a gradient of negative eight over three.

This inequality is, therefore, π¦
is greater than or equal to negative eight over three π₯ plus eight. All maxima and minima occur at the
corners of the region. Weβre specifically interested in
the maximum of the function. So by finding the value of π₯ and
π¦ at these points, we can determine which order pair when substituted into π
equals five π₯ plus two π¦ will give us the largest possible value for π.

Our vertices fall at seven, zero;
seven, eight; zero, eight; and three, zero. At zero, eight, π is equal to five
times zero, add two times eight, which gives us a value of 16. At three, zero, π is equal to five
times three, add two times zero, which is 15. At seven, eight, π is equal to
five times seven, add two times eight, which is 51. And at seven, zero, π is equal to
five times seven, add two times zero, which is 35.

Our function π equals five π₯ plus
two π¦ is therefore greatest when π₯ equals seven and π¦ equals eight. The values of π₯ and π¦ that
maximise our function are seven, eight.